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3 years ago

I need help with my geometry homework! Thank you! I will mark brainliest!!!!

Mathematics

1 answer:

Firlakuza [10]3 years ago

3 0

Answer:

The first choice is the one you want

Step-by-step explanation:

Use geometric means to solve for a first. The formula is

$8^2=15a$ and 64 = 15a and a = 64/15

That one was quite easy. Finding b is a bit more difficult.

The formula for finding b is

$\frac{a}{b}=\frac{b}{15+a}$

Solving for b:

$b^2=a(15+a)$

Filling in we have

$b^2=15(\frac{64}{15})+(\frac{64}{15})^2$ and

$b^2=64+\frac{4096}{225}$ and

$b^2=\frac{18496}{225}$ so b = 136/15

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Find the area of the region under the graph of f on [a, b]. f(x) = x2 − 8x + 17; [−1, 2]

Harman [31]

Answer:

The area is 42 area units.

Step-by-step explanation:

The area of a function f(x) on an interval [a,b] is given by:

$A = \int\limits^a_b {f(x)} \, dx$

Applying to this question:

$A = \int\limits^-1_2 {x^2 - 8x + 17} \, dx$

Then

$F(x) = \frac{x^{3}}{3} - 4x^{2} + 17x$

$A = F(2) - F(-1)$

$F(2) = frac{2^{3}}{3} - 4*2^{2} + 17*2 = \frac{8}{3} + 18 = \frac{8 + 3*18}{3} = \frac{62}[3}$

$F(-1) = frac{(-1)^{3}}{3} - 4*(-1)^{2} + 17*(-1) = -\frac{1}{3} - 21 = \frac{-1 - 21*3}{3} = -\frac{64}[3}$

$A = F(2) - F(-1) = \frac{62}[3} - (-\frac{64}[3}) = \frac{62+64}{3} = 42$

The area is 42 area units.

5 0

3 years ago

OAA', OBB', and OCC' are straight lines. Triangle ABC is mapped onto Triangle A'B'C' by an enlargement with center O. What is th

bazaltina [42]

Answer:

(D) 2

Step-by-step explanation:

The scale factor of the enlargement of ΔABC to ΔA'B'C' is given by the ratio of the length of the corresponding sides of ΔA'B'C' and ΔABC

Therefore, we have;

$The \ scale \ factor = \dfrac{Length \ of \overline {B'C'}}{Length \ of \overline {BC}} = \dfrac{Length \ of \overline {A'C'}}{Length \ of \overline {AC}} = \dfrac{Length \ of \overline {A'B'}}{Length \ of \overline {AB}}$

$\dfrac{Length \ of \overline {B'C'}}{Length \ of \overline {BC}} = \dfrac{2 \ units}{1 \ unit} = 2$

$\dfrac{Length \ of \overline {A'C'}}{Length \ of \overline {AC}} = \dfrac{4 \ units}{2 \ units} = 2$

$\dfrac{Length \ of \overline {A'B'}}{Length \ of \overline {AB}} = \dfrac{2 \cdot \sqrt{5} \ units}{\sqrt{5} \ units} = 2$

Therefore, the scale factor = 2

6 0

3 years ago

What is the value of p 13²=p

solniwko [45]

169 because 13 x 13 = 168

8 0

3 years ago

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8. Solve for x. Round your answer to the nearest tenth. 8.6 4 x 0176

earnstyle [38]

Answer:

C. 9.5

Step-by-step explanation:

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8 0

2 years ago

Continue the following pattern; describe the eighth figure in this pattern a. 49 small squares forming a larger 8 by 8 square b.

olga2289 [7]

Answer:

l think the answer is B

Step-by-step explanation:

The eighth patern should give the result 81

7 0

3 years ago