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3 years ago

Use the right triangular prism to answer the question below

Mathematics

2 answers:

Dahasolnce [82]3 years ago

6 0

Answer:

Just multiply the triangle 5x5 then the side pieces 6x5 then add and u got your answer

lara [203]3 years ago

4 0

5x5= and 5x6= (5x5=) For the top
(5x6=) answer for bottom

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What is the equation of a line that passes through the points (-3, 4) and (2,8)?

ziro4ka [17]

Given that the line passes through the two points (-3,4) and (2,8)

We need to determine the equation of the line.

<u>Slope:</u>

The slope of the line can be determined using the formula,

$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting the points (-3,4) and (2,8) in the above formula, we get;

$m=\frac{8-4}{2+3}$

$m=\frac{4}{5}$

Thus, the slope of the line is $m=\frac{4}{5}$

<u>Equation of the line:</u>

The equation of the line can be determined using the formula,

$y-y_1=m(x-x_1)$

Substituting the point (-3,4) and $m=\frac{4}{5}$ in the above formula, we have;

$y-4=\frac{4}{5}(x+3)$

$y-4=\frac{4}{5}x+\frac{12}{5}$

$y=\frac{4}{5}x+\frac{12}{5}+4$

$y=\frac{4}{5}x+\frac{32}{5}$

Thus, the equation of the line is $y=\frac{4}{5}x+\frac{32}{5}$

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3 years ago

10 POINTS AND BRAINLIEST! PLEASE ANSWER BOTH QUESTIONS!

jok3333 [9.3K]

Segment CF is parallel to segment BE because these segments are side by side and will have the same distance continuously between them. Therefore your answer would be,

Segment BE

∠AOF is the angle that is the supplementary angle to ∠FOD because these are two angles that sum up to 180°. hence, the answer is,

∠AOF

8 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

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3 years ago

Question 4(Multiple Choice Worth 1 points)

lianna [129]

Answer:

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2 years ago