Find the hcf of 299, 589 and 779​

Answer: $2.46

Step-by-step explanation:

First, let's find the unit price for both buckets!

A 1 gallon bucket of paint is $21.95. But how much will a 1 gallon bucket of paint cost if it had origianlly been 97.45 for 5 gallons of paint?

We need to divide 97.45 by 5, and the answer will be the unit price.

97.45 ÷ 5 = 19.49.

A 1 gallon bucket of paint that was originally priced as $97.45 for 5 gallons of paint si now $19.49 per bucket.

But we're not done! We still need to find the difference in unit prince between a 5 gallon bucket of paint for $97.45 and a 1 gallon bucket of paint for $21.95.

21.95 - 19.49 = 2.46.

Therefore, the difference in unit price between a 5 gallon bucket of paint for $97.45 and a 1 gallon bucket of paint for $21.95 is $2.46.

9514 1404 393

Answer:

  total cooling was 252 degrees

Step-by-step explanation:

  quantity = rate × time

The total temperature change is ...

  (-6°/h)(42 h) = -252°

The total change in temperature was a cooling of 252 degrees.

9514 1404 393

Answer:

  y = x² +x -6

Step-by-step explanation:

The graph shows the zeros of the function to be -3 and +2. This means the factored form of the function can be written as ...

  y = (x -(-3))(x -(+2)) = (x+3)(x-2)

Multiplying this out, we get the desired standard form:

  y = x² +x -6

We note that y=-6 when x=0, which matches the y-intercept. Nothing else needs to be done to this equation.

Answer:

1, 2, 3, 4

Step-by-step explanation:

x+(x+1)+(X+2)+(X+3)+2 = 3(X+3)

4X+8=3x+9

x=1

Answer:

5,6,7,8

Step-by-step explanation:

5+6+7+8=26

8*3=24

26-24=2

Answer:

answer should be 13

Step-by-step explanation:

look to photo

Answer:

The volume is increasing at a rate of 25,133 mm/s.

Step-by-step explanation:

Diameter is 40mm

Radius is half the diameter, so [tex]r = \frac{40}{2} = 20[/tex]

How fast is the volume increasing when the diameter is 40mm?

We have to apply implicit differentiation, of V and r in function of t. So

[tex]V = \frac{4}{3} \pi r^3[/tex]

[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]

The radius of a sphere is increasing at a rate of 5 mm/s.

This means that [tex]\frac{dr}{dt} = 5[/tex]

Then

[tex]\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}[/tex]

[tex]\frac{dV}{dt} = 4\pi (20)^2(5)[/tex]

[tex]\frac{dV}{dt} = 25133[/tex]

The volume is increasing at a rate of 25,133 mm/s.

Answer:

im not exactly sure how to do this one but i think that you can use pythagoras theorem which is a^2 +b^2=c^2. In the question you can use d^2 +e^2= f^2

Answer:

2

Step-by-step explanation:

18+16+4=36

40-38=2

Answer:

2 cupcakes left

Step-by-step explanation:

40 - (18 + 16 + 1 + 1 + 1 + 1) = 2

Answer:

34

Step-by-step explanation:

Answer:

Mia would have a 55$ interest  :)

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given:

Price of 12 shirts = Rs. [tex]3600\dfrac{3}{5}[/tex]

Price of 6 pants = Rs. [tex]3000\dfrac{1}{5}[/tex]

To find:

The price of 4 shirts and 4 pants​.

Solution:

It is given that,

Price of 12 shirts = Rs. [tex]3600\dfrac{3}{5}[/tex]

Price of 1 shirt = Rs. [tex]\dfrac{1}{12}\times 3600\dfrac{3}{5}[/tex]

Price of 4 shirts = Rs. [tex]\dfrac{4}{12}\times 3600\dfrac{3}{5}[/tex]

                          = Rs. [tex]\dfrac{1}{3}\times \dfrac{18003}{5}[/tex]

                          = Rs. [tex]\dfrac{18003}{15}[/tex]

Price of 6 pants = Rs. [tex]3000\dfrac{1}{5}[/tex]

Price of 1 pant = Rs. [tex]\dfrac{1}{6}\times 3000\dfrac{1}{5}[/tex]

Price of 4 pants = Rs. [tex]\dfrac{4}{6}\times 3000\dfrac{1}{5}[/tex]

                          = Rs. [tex]\dfrac{2}{3}\times \dfrac{15001}{5}[/tex]

                          = Rs. [tex]\dfrac{30002}{15}[/tex]

Now, the price of 4 shirts and 4 pants​ is:

[tex]\text{Required price}=\dfrac{18003}{15}+\dfrac{30002}{15}[/tex]

[tex]\text{Required price}=\dfrac{48005}{15}[/tex]

[tex]\text{Required price}=\dfrac{9601}{3}[/tex]

[tex]\text{Required price}=3200\dfrac{1}{3}[/tex]

Therefore, the price of 4 shirts and 4 pants​ is Rs. [tex]3200\dfrac{1}{3}[/tex].

Answer:

320.28?

Step-by-step explanation:

Answer:

Answer:

Step-by-step explanation:

take 60 degree as reference angle

using tan rule

tan 60=opposite/adjacent

[tex]\sqrt{3[/tex]=x/5

5*[tex]\sqrt{3[/tex] =x

5[tex]\sqrt{3[/tex]=x

Answer:

This question would be answered by doing devision because you are splitting  the amount of cookies that should be delivered amongst 13 girls.

1,000/13

76.92307692

That is what came out in my calculator. I don't know if you need it in decimal or in a solid integer but if you were to round 76.9 you would get 77, it is always good to check your work so,

77 times 13

1001

That equals 1001 so that would help at least reach their goal of 1,000 boxes of cookies.

Step-by-step explanation:

I really hope this helps!

Answer:

A. 15

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Trigonometry

Step-by-step explanation:

Step 1: Define

Identify variables

Angle θ = 37°

Opposite Leg = x

Hypotenuse = 25

Step 2: Solve for x

Step-by-step explanation:

The given expression is:

[tex]r=\dfrac{2p^2}{q-3}[/tex]

We need to solve it to make q the subject of the formula.

Cross multiplying both sides,

[tex]rq-3r=2p^2[/tex]

Adding 3r to both sides,

[tex]rq-3r+3r=2p^2+3r\\\\rq=2p^2+3r[/tex]

Now, dividing both sides by r,

[tex]q=\dfrac{2p^2+3r}{r}[/tex]

Hence, this is the required solution.

Answer:The value of log base 3 log base 2 log base √3 81 is -a)1 b) 2 c)3 d) 0

Answer:

4

Step-by-step explanation:

Define [tex]\log_a b=c\implies a^c=b[/tex].

Let

[tex]\log_3 81=x[/tex].

Using our definition [tex]\log_a b=c\implies a^c=b[/tex], we have:

[tex]3^x=81[/tex]

Solving for [tex]x[/tex]:

[tex]x=\boxed{4}\text{ from simply knowing that }3^4=81[/tex]

Or

Algebraically solve step-by-step by taking the log of both sides:

[tex]\log 3^x=\log 81[/tex]

Using log property [tex]\log a^b=b\log a[/tex], rewrite:

[tex]x\log 3=\log81[/tex]

Divide both sides by log(3):

[tex]x=\frac{\log 81}{\log 3}=\boxed{4}[/tex]

Answer:

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c[/tex]

Step-by-step explanation:

The question is not properly formatted. However, the integral of [tex]\int {7 \sec(\theta) } \, d\theta[/tex] is as follows:

[tex]\int {7 \sec(\theta) } \, d\theta[/tex]

Remove constant 7 out of the integrand

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta[/tex]

Multiply by 1

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta[/tex]

Express 1 as: [tex]\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta[/tex]

Expand

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta[/tex]

Let

[tex]u = \sec(\theta) + \tan(\theta)[/tex]

Differentiate

[tex]\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)[/tex]

Make [tex]d\theta[/tex] the subject

[tex]d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}[/tex]

So, we have:

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}[/tex]

Cancel out [tex]\sec(\theta)\tan(\theta) + sec^2(\theta)[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}[/tex]

Integrate

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c[/tex]

Recall that: [tex]u = \sec(\theta) + \tan(\theta)[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c[/tex]

Answer:

[tex]tom = \frac{85}{100} \times 20 \\ = 17 \: questions[/tex]

Melissa had a greater mark.

Answer:

Melissa

Step-by-step explanation:

Melissa: 18/20

Tom: 85/100

You need a common denomenater so multiply 18 and 20 by 5:

90/100>85/100

Melissa has the greater mark by 5%

Answer:

The null hypothesis is [tex]H_0: p \geq 0.32[/tex]

The alternative hypothesis is [tex]H_1: p < 0.32[/tex]

Step-by-step explanation:

State the null and alternative hypotheses to test that less than 32% of mall visitors make a purchase

At the null hypothesis, we test if the proportion is of at least 32%, that is:

[tex]H_0: p \geq 0.32[/tex]

We are testing if less than 32% of visitors make a purchase, so, at the alternative hypothesis, we test if the proportion is less than 32%, that is:

[tex]H_1: p < 0.32[/tex]

Answer:

A box plot is drawn with end points at 24 and 49.The box extends from 28 to 44 and a vertical line is drawn inside the box at 34.

Step-by-step explanation:

Ordering the data given :

24,28,32,34,40,44,49

We can calculate the 5 number summary required to give the appropriate boxplot that can be produced :

Minimum = 24

Maximum = 49

Median = 1/2(n+1)th term

n = 7

Median = 1/2(8) = 4th term

Median = 34

Lower quartile, Q1 = 1/4(n+1)th term

n = 7

1/4(8) = 2nd term

Q1 = 28

Upper quartile : 3/4(n+1)th term

n = 7

Q3 = 3/4(8) = 6th term

Q3= 44

The g(x) is equal to (2/3)x.To find g(x) in the composite function h(x) = f(g(x)), we need to substitute f(x) = 2x - 1 into the equation and solve for g(x

Given that h(x) = (4/3)x - 1 and f(x) = 2x - 1, we can set up the composite unction:

h(x) = f(g(x))

Substituting f(x) into the equation:

(4/3)x - 1 = 2(g(x)) - 1

Simplifying the equation:

(4/3)x - 1 = 2g(x) - 1

We can now solve for g(x):

2g(x) = (4/3)x

Dividing both sides by 2:

g(x) = (2/3)x

For more such questions on function

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Answer:

9.43

Step-by-step explanation:

Given

See attachment for illustration

Required

Distance AE

The actual path taken by the taxicab are in bold lines.

To calculate the distance AE, we simply consider the triangle AOE where

[tex]AE^2 = OE^2 + AO^2[/tex]

[tex]AE^2 = 5^2 + (5+3)^2[/tex]

[tex]AE^2 = 5^2 + 8^2[/tex]

[tex]AE^2 = 25 + 64[/tex]

[tex]AE^2 = 89[/tex]

Take square roots

[tex]AE = \sqrt{89[/tex]

[tex]AE = 9.43[/tex]

Pythagoras theorem states that for a right angle triangle the measure of the  square of the hypotenuse side is equal to the sum of the square of the other two sides.The ladder has to be kept at 13.3( round to the nearest tenth) feet away from the bottom of the wall.

Given information-

The length of the ladder is 24 foot.

The height of the wall is 20 foot.

Suppose the ladder has to kept at x feet away from the bottom of the wall

Here a right angle is formed in which the ladder is hypotenuse side shown in the diagram attached below.

Pythagoras theorem states that for a right angle triangle the measure of the  square of the hypotenuse side is equal to the sum of the square of the other two sides.

As the ladder has to be kept at x feet away from the bottom of the wall. Thus,

[tex]24^2=20^2+x^2\\ 576=400+x^2[/tex]

Solve for the x,

[tex]x=\sqrt{576-400}\\ x=\sqrt{176}\\ x=13.266[/tex]

hence the ladder ladder has to be kept at 13.3( round to the nearest tenth) feet away from the bottom of the wall.

Learn more about the Pythagoras theorem here;

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Answer:

el resultado es 8 ese es el resultado

Answer:

8

I used a calculator and got this answer so it should be right.

Answer:

720 km.

Step-by-step explanation:

120 * 6 = 720 km

Answer:

720

Step-by-step explanation:

120km every hour times by 6

Answer:

-2/7

Step-by-step explanation: