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3 years ago

6

What is the length of AC?

1 answer:

Alenkasestr [34]3 years ago

5 0

So in order to find line AC you must find line AD and DC then plus them together.

to find AD use Pythagoras theorem
a^2 = c^2 - b^2
AD^2 = 7.5^2 - 6.5^2
AD^2 = 56.25 - 42.25
AD^2 = 14
square root both sides to get rid of the ^2
AD ≈ 3.7 or 3.74

Do the same for DC
DC^2 = 10^2 - 6.5^2
DC^2 = 100 - 42.25
DC^2 = 57.75
DC ≈ 7.6

now plus AD and DC which should give u 11.3

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What is the surface area of the right prism below ?

topjm [15]

Answer:

Step-by-step explanation:

First find the area of the 2 bases, the triangles. 6x8= 48\2= 24 + 24= 48

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3 years ago

Find the slope of the line that goes through the points (-11, -19) & (-11, -10)

anyanavicka [17]

Answer:

<h2>The slope is undefined</h2>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are (-11, -19) & (-11, -10)

The slope is

$m = \frac{ - 10 - - 19}{ - 11 - - 11} = \frac{ - 10 + 19}{ - 11 + 11} = - \frac{9}{0} \\$

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Hope this helps you

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3 years ago

Write the function that describes a parabola that has a vertex at (3,-4) and goes through the point (5,-12).

Anna007 [38]

Answer: The answer is f(x) = - 2x²+12x-22.

Step-by-step explanation: We are given to write the function describing a parabola with vertex (3, -4) and passing through the point (5, -12).

We know that the standard form of a parabola with vertex (h, k) is given by

$f(x)-k=a(x-h)^2.$

Here (h, k) = (3, -4), so we have

$f(x)-(-4)=a(x-3)^2\\\\\Rightarrow f(x)+4=a(x-3)^2.~~~~~~~~~~~(I)$

Also, the parabola is passing through the point (5, -12), so

$-12+4=a(5-3)^2\\\\\Rightarrow -8=a\times 4\\\\\Rightarrow a=-2.$

Substituting the value of 'a' above in equation (I), we have

$f(x)+4=-2(x-3)^2\\\\\Rightarrow f(x)=-2(x^2-6x+9)-4\\\\\Rightarrow f(x)=-2x^2+12x-22.$

Thus, the answer is f(x) = - 2x²+12x-22..

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