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3 years ago

Hey ASAPPP PLS 80 points

Mathematics

2 answers:

Klio2033 [76]3 years ago

4 0

Answer:

using pythagoras theorem

a^2+b^2=c^2

8^2+x^2=(x+2)^2

64+x^2=x^2+2*x*2+2^2

64+x^2=x^2+4x+4(here x^2 of left hand side and x^2 of right hand side gets cancel)

64=4x+4

64-4=4x

60/4=x

15=x

Step-by-step explanation:

alexdok [17]3 years ago

4 0

Answer is up here, give them brainliest they did a great job ^^

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8. The graph of a horizontal line is a function. True False

11111nata11111 [884]

Answer:

It is not due to the fact it has no variable

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3 years ago

a force of 2 pounds is needed to stretch a spring 8 inches.what force is required to stretch the spring 16 inches

Korvikt [17]

Answer:

Four (4) pounds

Step-by-step explanation:

Because we required two (2) pounds to stretch the spring eight (8) inches, we need double the force to stretch it sixteen (16) inches. Sixteen is just double the original amount, meaning we need four (4) pounds to stretch the spring sixteen inches.

Hope this helps!

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3 years ago

The perimeter of a regular hexagon is ninety-six centimetres. Find the length of each side.

rusak2 [61]

Answer:

54cm, which is one side length

Step-by-step explanation:

The sum of all of the sides are 6, it's a regular hexagon the perimemter is just six times one side 36cm

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8 0

3 years ago

What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

maks197457 [2]

Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

$y=\dfrac{4}{2}$

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

3 0

2 years ago

Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)

Sphinxa [80]

Answer:

x=2.125

y=0

C=19.125

Step-by-step explanation:

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The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.

So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.

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2 years ago