PLS HELP :Find all the missing elements:

Answer: 93-(18+30)=g

93-48=g

45=g

Step-by-step explanation: yup

The answer is 93-18-30-g=0 or 18+30+g=93

Answer:

Hello,

Step-by-step explanation:

[tex]y^2-x^2+1=50\\y^2=x^2+49\\2\ functions \ :\\\\y=\sqrt{x^2+49} \\\\or\\\\y=-\sqrt{x^2+49} \\[/tex]

Using function concepts, it is found that:

----------------------

The expression is given by:

[tex]y^2 - x^2 + 1 = 50[/tex]

In terms of x, the equation is given by:

[tex]y^2 = 50 + x^2 - 1[/tex]

[tex]y^2 = x^2 + 49[/tex]

[tex]y = \pm \sqrt{x^2 + 49}[/tex]

----------------------

Testing the input x = 0:

[tex]y = \pm \sqrt{0^2 + 49}[/tex]

[tex]y = \pm \sqrt{49}[/tex]

[tex]y = \pm 7[/tex]

Two output values for one input, thus, it is not a function.

A similar problem is given at https://brainly.com/question/24603090

Answer:

see below

Step-by-step explanation:

-33, -27, -21, -15,....

-33 +6 = -27  

-27+6 = -21

-21+6 = -15

This is an arithmetic sequence

The common difference is +6

explicit formula

an=a1+(n-1)d  where n is the term number and d is the common difference

an = -33 + ( n-1) 6

an = -33 +6n -6

an = -39+6n

recursive formula

an+1 = an +6

10th term

n =10

a10  = -39+6*10

      = -39+60

      =21

sum formula

see image

The sum will diverge since we are adding infinite numbers

Answer: c = 4.97 and c = -1.97

Step-by-step explanation: Mean Value Theorem states if a function f(x) is continuous on interval [a,b] and differentiable on (a,b), there is at least one value c in the interval (a<c<b) such that:

[tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex]

So, for the function f(x) = [tex]2x^{3}-9x^{2}-60x+1[/tex] on interval [-4,9]

[tex]f'(x) = 6x^{2}-18x-60[/tex]

f(-4) = [tex]2.(-4)^{3}-9.(-4)^{2}-60.(-4)+1[/tex]

f(-4) = 113

f(9) = [tex]2.(9)^{3}-9.(9)^{2}-60.(9)+1[/tex]

f(9) = 100

Calculating average:

[tex]6c^{2}-18c-60 = \frac{100-113}{9-(-4)}[/tex]

[tex]6c^{2}-18c-60 = -1[/tex]

[tex]6c^{2}-18c-59 = 0[/tex]

Resolving through Bhaskara:

c = [tex]\frac{18+\sqrt{1740} }{12}[/tex]

c = [tex]\frac{18+41.71 }{12}[/tex] = 4.97

c = [tex]\frac{18-41.71 }{12}[/tex] = -1.97

Both values of c exist inside the interval [-4,9], so both values are mean slope: c = 4.97 and c = -1.97

Step-by-step explanation:

Exterior angle BOA = 250°

Interior angle BOA = 360°- 250° = 110°

Now,

(A) BDA = interior angle BOA / 2 = 55°( Property of circles)

(B) From the figure, we observe that AOBC is a cyclic quadrilateral (i.e. sum of opposite angles is 180°).

Therefore, BCA + BOA = 180°

BCA = 180° - 110° = 70°

Answer:

10 km

Step-by-step explanation:

Distance = 5 cm

4 cm = 8 km

In km, how far apart is school and home?

Cross Multiply

[tex]\frac{4cm}{8km}[/tex] · [tex]\frac{5cm}{1}[/tex]

Cancel centimeters

[tex]\frac{40(km)(cm)}{4cm}[/tex]

Divide

= [tex]\frac{40km}{4}[/tex]

= 10 km

Answer:

The first and second iteration of Newton's Method are 3 and [tex]\frac{11}{6}[/tex].

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form [tex]f(x) = 0[/tex] based on the following formula:

[tex]x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}[/tex]

Where:

[tex]x_{i}[/tex] - i-th Approximation, dimensionless.

[tex]x_{i+1}[/tex] - (i+1)-th Approximation, dimensionless.

[tex]f(x_{i})[/tex] - Function evaluated at i-th Approximation, dimensionless.

[tex]f'(x_{i})[/tex] - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be [tex]f(x) = x^{2}-8[/tex] and [tex]f'(x) = 2\cdot x[/tex], the resultant expression is:

[tex]x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}[/tex]

First iteration: ([tex]x_{1} = 2[/tex])

[tex]x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}[/tex]

[tex]x_{2} = 2 + \frac{4}{4}[/tex]

[tex]x_{2} = 3[/tex]

Second iteration: ([tex]x_{2} = 3[/tex])

[tex]x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}[/tex]

[tex]x_{3} = 2 - \frac{1}{6}[/tex]

[tex]x_{3} = \frac{11}{6}[/tex]

Answer:

(-5/9, 16/9)

Step-by-step explanation:

2y = -x + 3

-y = 5x + 1

To find the intersection, you need to substitute the y-value from the second equation into the first equation.  Rearrange the second equation so that it is equal to y.

-y = 5x + 1

-1(-y) = -1(5x + 1)

y = -5x - 1

Substitute this equation into the y-value of the first equation.

2y = -x + 3

2(-5x - 1) = -x + 3

-10x - 2 = -x + 3

(-10x - 2) + 2 = (-x + 3) + 2

-10x = -x + 5

(-10x) + x = (-x + 5) + x

-9x = 5

(-9x)/(-9) = (5)/(-9)

x = -5/9

Plug this x value into one of the equations and solve for y.

2y = -x + 3

2y = -(-5/9) + 3

2y = 5/9 + 3

2y = 32/9

(2y)/2 = (32/9)/2

y = 32/18 = 16/9

The ordered pair is (-5/9, 16/9).

Answer:

x= (2+ √ 34) /5 , (2- √ 34) /5

decimal form= 1.566

Step-by-step explanation:

Answer:

x=4

Step-by-step explanation:

To solve for x, we must get x by itself on one side of the equation.

[tex]\sqrt{x} +5-3=4[/tex]

First, we can combine like terms on the left side. Subtract 3 from 5.  

[tex]\sqrt{x} +(5-3)=4[/tex]

[tex]\sqrt{x} +2=4[/tex]

2 is being added on to the square root of x. The inverse of addition is subtraction. Subtract 2 from both sides of the equation.

[tex]\sqrt{x} +2-2=4-2[/tex]

[tex]\sqrt{x} = 4-2[/tex]

[tex]\sqrt{x} =2[/tex]

The square root of x is being taken. The inverse of a square root is a square. Square both sides of the equation.

[tex](\sqrt{x} )^{2} =2^2[/tex]

[tex]x=2^2[/tex]

Evaluate the exponent.

2^2= 2*2 =4

[tex]x=4[/tex]

Let's check our solution. Plug 4 in for x and solve.

[tex]\sqrt{x} +5-3=4[/tex]

[tex]\sqrt{4} +5-3=4[/tex]

[tex]2+5-3=4[/tex]

[tex]7-3=4[/tex]

[tex]4=4[/tex]

Our solution checks out, so we know x=4 is correct.

Answer:

46

Step-by-step explanation:

See attached for reference

The points given:

They form a rectangle as seen in the picture.

We can notice that this is a parallelogram, as respective lines have same difference of coordinates.

So calculating only the two of the sides will be sufficient to get its perimeter:

So, the perimeter:

Answer:

4th multiple = 24

Step-by-step explanation:

Given

Let the two numbers be represented by m and n

Required

Find the 4th common multiple of the numbers.

From the question, we understand that the first common multiple of m and n is 6.

This can be represented as:

m * n * 1 = 6

mn = 6

Their fourth common multiple can be represented as: m * n * 4

4th multiple = m * n * 4

4th multiple = 4 * mn

Substitute 6 for mn

4th multiple = 4 * 6

4th multiple = 24

Hence, the 4th multiple of both numbers is 24.

Answer:

2x-1=3

2x=3+1

2x=4

x=2

[tex]\huge\boxed{\underline{\bf \: Answer}}[/tex]

(a) Let's try with x = - 1

[tex] \sf \: 2x - 1 = 3 \\\sf 2( - 1) - 1 = 3 \\ \sf- 2 - 1 = 3 \\ \\ \boxed{\bf- 3 \: \bcancel= \: 3}[/tex]

So, x = - 1 is not the solution to the given equation.

______________

(b) Now, try with x = 2

[tex]\sf2x - 1 = 3 \\ \sf2(2) - 1 = 3 \\ \sf4 - 1 = 3 \\ \\ \boxed{\bf3 = 3}[/tex]

Yes, we can see that x = 2 is the correct solution for the equation.

______________

Hope it helps.

RainbowSalt2222

Answer:

(-25)+(-12)+(-34) = -71

so when you add negative numbers you simply add them such as -2+-2 -4

so same conditions

so it will be -25+-12+-34 and it will simply be 25+12+34 so -71

Answer:

Step-by-step explanation:

Given that:

[tex]T(x,y) = \dfrac{100}{1+x^2+y^2}[/tex]

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

[tex]c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)[/tex]

By cross multiply

[tex]c({1+x^2+2y^2}) = 100[/tex]

[tex]1+x^2+2y^2 = \dfrac{100}{c}[/tex]

[tex]x^2+2y^2 = \dfrac{100}{c} - 1 \ \ -- (2)[/tex]

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

[tex]\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1[/tex]

This is the equation for the  family of the eclipses centred at (0,0) is :

[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]

[tex]a^2 = \dfrac{100}{c} -1 \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}[/tex]

Therefore; the level of the curves are all the eclipses with the major axis:

[tex]a = \sqrt{\dfrac{100 }{c}-1}[/tex]  and a minor axis [tex]b = \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}[/tex]  which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

Answer:

8 home games and 10 away games

Step-by-step explanation:

Total home goals

= 8+5+9+8

= 30

Number of home games

= 30/3.75

= 8

Total away game goals

= 7+8+4+5

= 24

Number of away games

= 24/2.4

= 10

Answer:

i think it is 8 home and 10 away matches

Step-by-step explanation:

should be 2 3 andd 5

think of the - (- as a plus sign (this is what i was always taught) to add them so it would in turn be (-5) + 12 which equals 7 and choice 3 and 5 also equal this

Answer:

C. Line chart

Step-by-step explanation:

Answer:

B. Histogram

Step-by-step explanation:

Histogram uses time.

Answer:

I believe it is 12,393.86

Answer:

76.8 cm

Step-by-step explanation:

Answer:

76.8 cm

Step-by-step explanation:

38.4 cm + 38.4 cm = 76.8 cm

Answer:

13.82%

Step-by-step explanation:

Here we have proportion of U.S. full time college students= p = 0.60, Random sample = n = 6

Here we apply Binomial distribution .

p ( X =x ) = nCx * px * ( 1 -p) n-x

a) Exactly two.

P ( x = 2 ) =  6C2 * 0.602 * ( 1 -0.60) 6-2

= 0.1382

Step-by-step explanation:

mean add upp all the numbers and divide by how many they are

[tex]S_n=\dfrac{n(a_1+a_n)}{2}\\a_1=200\\a_n=1998\\n=?\\\\a_n=a_1+(n-1)d\\d=2\\1998=200+(n-1)\cdot2\\2n-2=1798\\2n=1800\\n=900\\\\S_{900}=\dfrac{900\cdot(200+1998)}{2}=450\cdot 2198=989100[/tex]

The Sum is 989100.

The sum of even numbers formula is determined by using the formula to find the arithmetic progression. The sum of even numbers goes on until infinity. The sum of even numbers formula can also be evaluated using the sum of natural numbers formula. We need to obtain the formula for 2 + 4+ 6+ 8+ 10 +...... 2n.

The sum of even numbers = 2(1 + 2+ 3+ .....n). This implies 2(sum of n natural numbers) = 2[n(n+1)]/2 = n(n+1)

Given:

a1 = 200

an= 1998

So, using formula

S= n(a1 + an)/2

now,

d=2

an= a1+(n-1)d

1998= 200 + (n-1) 2

1998-200= (n-1)2

1798/2=n-1

n= 900

S900= 900( 200 + 1998)/2

        =450*2198

       = 989100

Hence, the sum is 989100.

Learn more about this concept here:

https://brainly.com/question/18837188

#SPJ2

The answer is "[tex]\bold{\frac{2}{105}}[/tex]", and the further calculation can be defined as follows:

When the "r" is the greatest common divisor for the two fractions.

So, we will use Euclid's algorithm:  

[tex]\to \bold{(\frac{8}{15}) -(\frac{188}{35})}\\\\\to \bold{(\frac{8}{15} -\frac{188}{35})}\\\\\to \bold{(\frac{56-54}{105})}\\\\\to \bold{(\frac{2}{105})}\\\\[/tex]

this is  [tex]\bold{(\frac{8}{15}) \ \ mod \ \ (\frac{18}{35})}[/tex]

we can conclude that the GCD for [tex]\bold{\frac{54}{105}}[/tex], when divided by [tex]\bold{\frac{2}{105}}[/tex], will be the remainder is 0.  Rational numbers go from [tex]\bold{\frac{2}{105}}[/tex] with the latter being the highest.

So, the final answer is "[tex]\bold{\frac{2}{105}}[/tex]".

Learn more:

greatest rational number:brainly.com/question/16660879

Answer:

b = 10.5

Step-by-step explanation:

2(b-9) = 3

then:

2*b + 2*-9 = 3

2b - 18 = 3

2b = 3 + 18

2b = 21

b = 21/2

b = 10.5

check:

2(10.5 - 9) = 3

2*1.5 = 3

Answer:

(3) y+2=2(x-8)

Step-by-step explanation:

Substitute the point into the equation and see if it is true

(8,-2)

(1) y+2=2(x+8)

    -2+2 = 2(8+8)

    0 = 2(16)

False

(2) y-2=2(x-8)

  -2-2 = 2(8-8)

  -4 =2 (0)

  False

(3) y+2=2(x-8)

  -2+2 = 2( 8-8)

  0 = 2(0)

 True

(4) y-2=2(x+8)

  -2-2 = 2(8+8)

  -4 = 2(16)

False

Answer:

[tex]\boxed{y+2=2(x-8) }[/tex]

Step-by-step explanation:

[tex]x=8[/tex]

[tex]y=-2[/tex]

[tex]\sf Check \ the \ third \ option.[/tex]

[tex]-2+2=2(8-8)[/tex]

[tex]\sf Both \ sides \ must \ be \ equal.[/tex]

[tex]0=2(0)[/tex]

[tex]0=0[/tex]

Answer:

3 hours

Step-by-step explanation:

Downstream, the speeds add up:

It will take:

To ride 87 km.

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]