Question 1 (2 points) ✓ Saved
Emilio has a bank account that pays 4% interest each year. He has $200 in his
account. How much interest will he earn after 1 year?
$10
$6
$4
$8
2.10

Answer:

F(x)=-x^2

Step-by-step explanation:

The one example of function with range of only negative number is:

F(x)=-x^2

Why this? We know for every x, x^2 is positive. So -x^2 will be negative.

Answer:

Two statement are true.

O The function is increasing on the interval (-∞, 0).

O The function is increasing on the interval (0, ∞).

Step-by-step explanation:

The function h(x) = 3x is a linear function (straight line) with a slope of 3, and a y-intercept of 0.  The constant, positive slope means that h(x) will increase over the interval (-∞, ∞).

See the attached graph.

Answer:

y=-3.5(x−5)^2+17

Step-by-step explanation:

Let's assume that since this equation has a vertex that it is a parabola with the form y = ax^2 + bx + c.

Let's rewrite it in vertex form as:

y=a(x−h)^2+k

where h and k are the horizontal (h) and vertical (k) coordinates of the vertex.

The vertex of (5,17) gives us both h (5) and k (17),  Substituting these values gives us:

  y=a(x−5)^2+17

To find a, use the one given point the parabola intersects (11,-109) and solve for a:

y=a(x−5)^2+17

-109=a((11)−5)^2+17

-109=a(6)^2+17

-109-17=a(36)

36a = -126

a = -3.5

This leads us to:

y=-3.5(x−5)^2+17

See the attached graph.

There is not enough paint for all the canvases. 26.7 cm² of more white paint is required.

Area of each canvas = Length × width

Area of each canvas = 5.3 × 8.5

Performing multiplication

Area of each canvas = 45.05 cm²

Total area of canvas in the class = 45.05 × 30

Performing multiplication

Total area of canvas in the class = 1351.5 cm²

Available amount of white paint = 6 × 220.8

Performing multiplication

Available amount of white paint = 1324.8 cm²

The available amount is less than canvas area in the class.

Amount of paint required = 1351.5 - 1324.8

Performing subtraction

Amount of paint required = 26.7 cm²

Thus, 26.7 cm² of white paint is required.

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By cancelative property, the roots of quadratic equation x² + 4 · x - 20 are: Exact: - 2 + 2√6, - 2 - 2√6, Approximate: 2.899, -6.899.

In this problem we must factorize a quadratic equation and use an algebraic property known as cancelative property, defined as:

If a · b = 0, then a = 0 or b = 0.

In addition, we can apply the following factorization rule:

x² - (r₁ + r₂) · x + r₁ · r₂ = (x - r₁) · (x - r₂)

First, factorize the quadratic equation:

x² - (- 2 + 2√6 - 2 - 2√6) · x + (- 2 + 2√6) · (- 2 - 2√6)

(x + 2 - 2√6) · (x + 2 + 2√6)

Second, apply cancelative property:

x + 2 - 2√6 = 0

x = - 2 + 2√6

x + 2 + 2√6 = 0

x = - 2 - 2√6

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A graph of the solutions of this inequality 3x + 2 ≥ 1/2(x + 5) is shown on the number line attached below.

In this exercise, we would solve the given inequality by making x the subject of formula as follows;

3x + 2 ≥ 1/2(x + 5)

By multiplying both sides of the inequality by 2, we have the following:

(3x + 2) × 2 ≥ 1/2(x + 5) × 2

6x + 4 ≥ x + 5

By rearranging and collecting like terms, we have the following:

6x - x ≥ 5 - 4

5x ≥ 1

x ≥ 1/5

Next, we would use an online graphing calculator to plot the inequality as shown in the graph attached below. Additionally, the circle on the number line (graph) is closed because of the greater than or equal to (≤) symbol.

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Answer:

24 days = 1500 tables

25 days = Fewer than 1500 tables

Step-by-step explanation:

So we have the inequality of < which is fewer than. Let's start by labeling all the numbers given.

What we want to find is how many days, so we'll make days = x

2100 is the initial and they sell 25 a day... 25 a x.

Combining them all we get a basic inequality of:

2100 - 25x < 1500

Let's get X alone.

Subtracting 2100 from both sides gives you

-25x < -600.

We have negatives on both sides. For me, if I see that, I make both of them positive;

25x < 600

Last step is to divide the 25 out of the X which means you'll divide 25 by 600 also

X = 24

Which means day 24 will be exactly 1500. We wanted fewer though, so adding a day gives you an answer of 25 days.

Answer:

d. Yes, this function is linear; the slope is -100 feet/second.

Step-by-step explanation:

Find the difference in height at two different time periods.

If

Δh = difference in height values

Δt = difference in seconds

then

[tex]\mathrm{slope = \dfrac{\Delta h}{\Delta y}}[/tex]

Take the difference in heights at t = 0 and t = 5

The height drops from 4000 to 3500 feet so Δh = 3500 - 400 = -500

Δt = 5- 0 = 5 seconds

slope = -500/5 = -100

If you compute the slope between any pair of heights and time difference you will get the same value

So slope is constant indicating a linear function

Answer:

$1,684.25

Step-by-step explanation:

(.85 x 1250) = 1062.50  If we take off 15% we leave on 85% which as a decimal is .85

Now add the tax

1062.50 + .06(1062.50)

1620.50 + 63.75

1684.25

Using the constant of proportionality the value for B in the table is obtained as option C: B = 40.

What is a COP?

The ratio between two quantities that are directly proportional is the constant of proportionality. When two quantities grow and shrink at the same rate, they are directly proportional.

The table contains three columns A, C, D.

The first row contains all the part values - 14, 28, 63

The second row contains all the whole values - 20, B, 90

It is necessary to calculate the constant of proportionality.

In this case, for A, we have 14 part and 20 whole, the constant will be -

COP = Whole / Part

COP = 20 / 14

COP = 1.42857

In this case, for C, we have 63 part and 90 whole, the constant will be -

COP = Whole / Part

COP = 90 / 63

COP = 1.42857

Since, the constant of proportionality is same for A and C, it will be same for B also.

In this case, for B, we have 28 part and B whole, the constant will be -

COP = Whole / Part

1.42857 = B / 28

B = 1.42857 × 28

B = 40

Therefore, the value for B is obtained as 40.

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A three-column table is given. Part A C D Part 14 28 63 Whole 20 B 90 What is the value of B in the table? 64 50 40 12

The equation has one solution

The solution to the equation is  x = 3

4(2x – 5) = 4

To solve this, we would first have to open the bracket.

4 * 2x - 4*5 = 4

8x - 20 = 4

We would want to have the variable x stand on its own

so we take like terms

8x = 4 + 20

8x = 24

To get the value of x divide through the equation by 8

x = 24 / 8

x = 3

The solution is just one at x = 3

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The resultant of the given expression (x² - 5x) (2x²+x-3) is 2x^4 - 7x^3 -8 x^2 + 15x .

What is Algebraic expression ?

Algebraic expression can be defined as the combination of variables and constants.

Given expression,

(x² - 5x) (2x²+x-3)

= x^2 ( 2x^2+x-3) - 5x(2x^2 + x - 3)

= 2x^4 + x^3 - 3x^2 - 5x * 2x^2 + x*-5x - 3 * -5x

= 2x^4 + x^3 -3x^2  - 10 x^3 - 5x^2 + 15x

= 2x^4 + x^3 - 10x^3 -3x^2 - 5x^2 + 15x

= 2x^4 - 7x^3 -8 x^2 + 15x .

Hence, The resultant of the given expression (x² - 5x) (2x²+x-3) is 2x^4 - 7x^3 -8 x^2 + 15x .

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The variable 'time to destination' quantitative.

Quantitative variables are those whose values are obtained through measurement or counting. A qualitative variable, also referred to as a category variable, is a non-numerical variable. It describes information that can be categorized. Examples like Eye colors, States, Dog breeds etc.

Any variable that isn't numerical is referred to as a qualitative variable or a category variable. It describes information that falls into a category. Generally speaking, a variable is considered to be quantitative if it can be used mathematically (such as addition). Otherwise, it's top-notch. You cannot, for instance, add blue and green.

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In the expression 2t+8, 8 represents the number of seats at each table.

An expression is combination of variables, numbers and operators.

Given that A coffee shop has a group of seats at the counter and t individual tables.

where each able has the same number of seats.

2t + 8 represents the total number of seats in the restaurant.

8 represents the number of seats at each table and 2t is the number of individual seats.

Hence, in the expression 2t+8, 8 represents the number of seats at each table.

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The area of the tin box without the top is 39100 square cm.

The cost of making this tin box is  Rs. 19550.

A cuboid is a three-dimensional closed figure which has volume along with surface area.

The volume of a cuboid is the product of its length, width, and height.

The total surface area of a cuboid is 2(lw + wh + wl) and the lateral surface area is 2(l + w)×h.

Given, A tin box of 1 m in length, 85cm in width, and 60cm in depth is to be made.

We know the total surface area of a cuboid is 2(lw + wh + wl).

Therefore, The surface area of the cuboid without the top would be,

= 2(lw + wh + wl) - (l×w).

= 2(100×85 + 85×60 + 85×100) - (85×60).

= 2(22100) - 5100.

= 44200 - 5100.

= 39100 square cm.

The cost of making this tin box is Rs.(39100×0.50).

= Rs. 19550.

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An expression to represent the number of sheets Alexia used is y = 3x + 40.

In math, expressions are mathematical statements with a minimum of two terms containing numbers or variables, or both, connected by an operator in between.

The mathematical operators can be any of addition, subtraction, multiplication, or division. For example, p + q is an expression where p and q are terms having an addition operator in between.

There are two types of expressions, numerical expressions - which contain only numbers, and algebraic expressions- which contain both numbers and variables.

let

x = the number of sheets per ream

y = the total number of sheets

Alexia used three reams of paper, so the total number of sheets per ream will be 3x since each ream has the same number of sheets.

After that, she used another 40 sheets. Therefore the expression will become as

y = 3x + 40 represents the total number of sheets Alexia used.

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[tex]\frac{3x^{2} }{x^{2} +x-6}[/tex] is the product of the rational expressions.

What are rational expressions?

Two polynomials' ratio is shown through rational expressions. It denotes a polynomial in the denominator as well as the numerator. It is an algebraic expression that contains an unknown variable and is similar to a fraction in that it is a ratio.

We are given [tex]\frac{3x^{2} }{x^{2} +x-6}[/tex]

This means that the expression has to be multiplied by the number 1 for obtaining the product of the rational expression.

We know that when anything is multiplied by the number 1, we get the number itself.

Hence, [tex]\frac{3x^{2} }{x^{2} +x-6}[/tex] is the product of the rational expressions.

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The probability of  4 in the tens place is [tex]\frac{1}{9}[/tex], the probability of  at least one 4 in the tens place or the units place is [tex]\frac{1}{5}[/tex], the probability no 4 in either place is [tex]\frac{4}{5}[/tex].

(a) total number of 2-digit terms =(99 - 10) + 1 = 90

n(s) = 90

A probability is a number that reflects the chance or likelihood that a particular event will occur. Probabilities can be expressed as proportions that range from 0 to 1, and they can also be expressed as percentages ranging from 0% to 100%

number of 2 digits that have 4 in tens place = 40, 41, 42 ......49

n(a) = 10

P(A) =[tex]\frac{ n(a)}{n(s)}[/tex]

Substituting the values

P(A) = [tex]\frac{10}{90}[/tex]

P(A) = [tex]\frac{1}{9}[/tex]

Therefore, P(A) = [tex]\frac{1}{9}[/tex]

(b) From the previous one, we know the P(A) = [tex]\frac{1}{9}[/tex]

Number of 4's in units place = 14, 24, 34.....94 = 9

n(b) = 9

P(B) = [tex]\frac{ n(b)}{n(s)}[/tex]

Substituting the values

P(B) = [tex]\frac{9}{90}[/tex]

P(B) =[tex]\frac{1}{10}[/tex]

P(A and B) = [tex]\frac{1}{10}[/tex]

Using OR probability, we have

P(A or B) = P(A) + P(B) - P(A and B)

P([tex]\frac{1}{9}[/tex] or [tex]\frac{1}{10}[/tex]) = [tex]\frac{1}{9}[/tex] + [tex]\frac{1}{10}[/tex] - [tex]\frac{1}{90}[/tex]

P([tex]\frac{1}{9}[/tex]or [tex]\frac{1}{10}[/tex]) = [tex]\frac{1}{5}[/tex]

Therefore, P([tex]\frac{1}{9}[/tex]or [tex]\frac{1}{10}[/tex]) = [tex]\frac{1}{5}[/tex]

(c) Probability of no 4 in either place = 1 - ( probability of 4 in either place)

= 1 - [tex]\frac{1}{5}[/tex] = [tex]\frac{4}{5}[/tex]

Therefore, probability of no 4 in either place = [tex]\frac{4}{5}[/tex]

Therefore, the probability that the integer chosen has a 4 in the tens place is [tex]\frac{1}{9}[/tex], P([tex]\frac{1}{9}[/tex] or [tex]\frac{1}{10}[/tex]) = [tex]\frac{1}{5}[/tex], and probability of no 4 in either place = [tex]\frac{4}{5}[/tex].

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The result of the division of the number by 100 gives 3.05109 in decimal.

A decimal is really a number that is divided into two parts: another whole and a fraction.

For the given number.

= 305.109 ÷ 100

Then point will shift to 2 decimals towards left.

= 3.05109

Thus, the result of the division of the number by 100 gives 3.05109 in decimal.

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The surface area of the rectangular prism is 112 square metes.

The sum of all area of each surface that make up an object is referred to as its surface area.

So that in the given question,

Area of a rectangle = length x width

i. area of its base = length x width

                            = 8 x 2

                            = 16 sq. m

ii. area of its front surface = length x width

                                      = 4 x 8

                                      = 32 sq. m

iii. area of it one of its sides = length x width

                                    = 4 x 2

                                    = 8 sq. m

Therefore,

surface area of the rectangular prism = (2 x 16) + (2 x 32) + (2 x 8)

                                       = 32 + 64 + 16

                                       = 112

The surface area of the rectangular prism is 112 square meters.

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To graph the algebraic rule y = 3x - 11, we need to consider input values ​​from -5 to 5 and also choose appropriate scaling for the x and y axes.

A common way to choose a scale is to use 1 or 2 increments on the x and y axes. This makes it easy to draw points and read charts easily.

The x-axis goes from -5 to 5, so you can scale from -5 -4 -3 -2 -1 0 1 2 3 4 5.

For the y-axis, I need to find the minimum and maximum values ​​of y given the input x values.

Substituting the x value into the formula y = 3x - 11 gives the y value.

y = 3(-5) - 11 = -16

y = 3(5) - 11 = 14

So for the y axis you can take -20 -15 -10 -5 0 5 10 15 20

So a good scale for the x-axis is -5 to 5 in increments of 1, and -20 to 20 in increments of 5 for the y-axis. Graphing the equation y = 3x - 11 using these scales makes the graph easier to read and understand.

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For the given similar triangles, the value of x = 5.

Similar triangles share the same associated angle measurements and proportional side lengths.

Due to their similar shapes, two figures that are not always the same size can appear comparable. We might claim, for instance, that all circles are equivalent. Squares and equilateral triangles share similar characteristics. Although similar figures need not be congruent in order to be comparable, congruent figures are always comparable.

Despite having the same shape, similar triangles have different diameters. In identical triangles, corresponding angles are equal. In analogous triangles, the ratio of the corresponding sides is the same. Any pair of equivalent sides' squares and any comparable triangle's area have the same ratio.

As the given triangles are similar,

the sides are in same ratio,

⇒ 56 / 8x - 5 = 24 / 15

⇒ 8x - 5 = 35

⇒ 8x = 40

⇒ x = 5

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The matching of the inequality with the statement will be 1-B, 2-D, 3-A, and 4-C.

Inequality is defined as an equation that does not contain an equal sign. Inequality is a term that describes a statement's relative size and can be used to compare these two claims.

The matching of inequality with the statement is given as,

The matching of the inequality with the statement will be 1-B, 2-D, 3-A, and 4-C.

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The roots of quadratic equations:

In this question we need to determine sixteen cases of quadratic equations, whose roots should be determined. This can be done by following procedure:

Now we proceed to show the solution to each case:

Case 1

b² - 4 = 0

b² = 4

b = ± 4

Case 2

4 · a² - 1 = 0

4 · a² = 1

a² = 1 / 4

a = ± 1 / 2

Case 3

2 · x² = 21 + 11 · x

2 · x² - 11 · x - 21 = 0

2 · [x² - (11 / 2) · x - 21 / 2] = 0

2 · [x² - (11 / 2) · x + 121 / 16] = 2 · (121 / 16) + 2 · (21 / 2)

2 · (x - 11 / 4)² = 289 / 8

(x - 11 / 4)² = 289 / 16

x - 11 / 4 = ± 17 / 4

x = 11 / 4 ± 17 / 4

Case 4

2 · n² = 32

n² = 16

n = ± 4

Case 5

5 · x² = 3 · x + 92

5 · x² - 3 · x - 92 = 0

5 · [x² - (3 / 2) · x - 92 / 5] = 0

5 · [x² - (3 / 2) · x] = 5 · (92 / 5)

5 · [x² - (3 / 2) · x + 9 / 16] = 5 · (92 / 5 + 9 / 16)

5 · (x - 3 / 4)² = 1517 / 16

(x - 3 / 4)² = 1517 / 80

x - 3 / 4 = ± √(1517 / 80)

x = 3 / 4 ± √(1517 / 80)

Case 6

4 · x² = 7 · x + 92

4 · x² - 7 · x - 92 = 0

4 · [x² - (7 / 4) · x - 23] = 0

4 · [x² - (7 / 4) · x + 49 / 64] = 92 + 49 / 64

4 · (x - 7 / 8)² = 5937 / 64

x - 7 / 8 = ± √(5937 / 256)

x = 7 / 8 ± √(5937 / 256)

Case 7

5 · k² = - 7 · k + 138

5 · k² + 7 · k - 138 = 0

5 · [k² + (7 / 5) · k - 138 / 5] = 0

5 · [k² + (7 / 5) · k] = 5 · (138 / 5)

5 · [k² + (7 / 5) · k + 49 / 100] = 5 · (138 / 5 + 49 / 100)

5 · (k + 7 / 10)² = 2809 / 20

k + 7 / 10 = ± √(2809 / 100)

k = - 7 / 10 ± √(2809 / 100)

Case 8

m² - 88 = 3 · m

m² - 3 · m - 88 = 0

m² - 3 · m - 352 / 4 = 0

m² - 3 · m = 352 / 4

m² - 3 · m + 9 / 4 = 361 / 4

(m - 3 / 2)² = 361 / 4

m - 3 / 2 = ± √(361 / 4)

m = 3 / 2 ± √(361 / 4)

Case 9

4 · n² - 7 = 12 · n

4 · n² - 12 · n = 7

4 · (n² - 3 · n + 9 / 4) = 7 + 9

4 · (n - 3 / 2)² = 16

n - 3 / 2 = ± 4

n = 3 / 2 ± 4

Case 10

6 · n² - 2 · n = 140

6 · [n² - (1 / 3) · n] = 140

6 · [n² - (1 / 3) · n + 1 / 36] = 140 + 1 / 6

6 · (n - 1 / 6)² = 841 / 6

n - 1 / 6 = ± 841 / 36

n = 1 / 6 ± 841 / 36

Case 11

4 · p² = 64

p² = 16

p = ± 4

Case 12

6 · v² - 10 · v = 84

6 · [v² - (5 / 3) · v] = 84

6 · [v² - (5 / 3) · v + 25 / 36] = 84

6 · (v - 5 / 6)² = 84

(v - 5 / 6)² = 14

v - 5 / 6 = ± 14

v = 5 / 6 ± 14

Case 13

3 · r² - 77 = - 10 · r

3 · r² + 10 · r = 77

3 · [r² + (10 / 3) · r] = 77

3 · [r² + (10 / 3) · r + 25 / 9] = 77 + 25 / 9

3 · (r + 5 / 3)² = 718 / 9

(r + 5 / 3)² = 718 / 27

r + 5 / 3 = ± √(718 / 27)

r = - 5 / 3 ± √(718 / 27)

Case 14

5 · x² - 11 · x = - 6

5 · [x² - (11 / 5) · x + 121 / 100] = - 6 + 5 · (121 / 100)

5 · (x - 11 / 10)² = 1 / 20

(x - 11 / 10)² = 1 / 100

x - 11 / 10 = ± √(1 / 10)

x = 11 / 10 ± √(1 / 10)

Case 15

2 · n² = 11 · n + 6

2 · n² - 11 · n = 6

2 · [n² - (11 / 2) · n] = 6

2 · [n² - (11 / 2) · n + 121 / 16] = 6 + 2 · (121 / 16)

2 · (n - 11 / 4)² = 169 / 8

(n - 11 / 4)² = 169 / 16

n - 11 / 4 = ± 13 / 4

n = 11 / 4 ± 13 / 4

Case 16

4 · x² = 100

x² = 25

x = ± 5

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The equation for the circle with a diameter whose endpoints are (2, -5) and (-3, 1) is (x - 1)^2 + (y - 1)^2 = 13.

The endpoints are (2, -5) and (-3,1).

The standard form equation for a circle is known as

[tex](x-a)^2+(y-b)^2=r^2[/tex]

where (a, b) are the center's coordinates and (r) is the radius.

Taking into account the stated endpoints of the diameter. The center will then be at the midpoint, and the radius will be the distance between the center and either of the two endpoints.

Calculating the midpoint is as follows:

[tex]\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)[/tex]

where, [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] are two points.

Now putting the values

[tex]\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{-2+4}{2}, \frac{3-1}{2}\right)[/tex]

[tex]\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{2}{2}, \frac{2}{2}\right)[/tex]

[tex]\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(2, 1)[/tex]

Now we determine the equation of circle.

The center (2, 1) and the terminal are the two points (-2, 3).

r = [tex]\sqrt{(-2-1)^2+(3-1)^2}[/tex]

r =[tex]\sqrt{(-3)^2+(2)^2}[/tex]

r = [tex]\sqrt{9+4}[/tex]

r = √13

Now we can write the equation of circle as:

(x - 1)^2 + (y - 1)^2 = (√13)^2

(x - 1)^2 + (y - 1)^2 = 13

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The complete question is:

Find the equation for the circle with a diameter whose endpoints are (2, -5) and (-3, 1).

Write the standard equation for the circle.

The solution to the system of equations in this problem is given as follows:

(2,8).

The system of equations for this problem is defined as follows:

To eliminate the variable y, we multiply the first equation by 3 and the second by 5, hence:

Adding the two equations, the solution for x is obtained as follows:

47x = 94

x = 94/47

x = 2.

Hence the solution for y is obtained as follows:

4(2) + 5y = 48

5y = 40

y = 40/5

y = 8.

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