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3 years ago

7

HELPP I WILL GIVE BRAINLEST AND BUNCH OF POINTS

2 answers:

Kryger [21]3 years ago

3 0

Use the formulas for the area of a triangle and the area of a rectangle for both figures. The bases are the tops and bottoms of the figures and the lateral sides are the sides not including the top or bottom. It’s basically just finding the areas of each side and adding them all up to find the surface area. Volume is finding the area of the base, using the triangle area formula, and multiplying that by the height of the figure
Hope this helps!

Bogdan [553]3 years ago

3 0

Answer:

Step-by-step explanation:

Use the formulas for the area of a triangle and the area of a rectangle for both figures. The bases are the tops and bottoms of the figures and the lateral sides are the sides not including the top or bottom. It’s basically just finding the areas of each side and adding them all up to find the surface area. Volume is finding the area of the base, using the triangle area formula, and multiplying that by the height of the figure

Hope this helps!

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3 years ago

if a line contains a the point (0,-1) and has a slope of 2 which of the following points also lies on the line

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3 years ago

M is the MIDPOINT OF FG. FG = 26 AND GM = (3X - 2)

vova2212 [387]

It is given in the question that

M is the MIDPOINT OF FG.

$FG = 26 \ AND \ GM = (3X - 2)$

Since M is the midpoint of FG, therefore GM is half of FG. That is

$3X-2 = \frac{1}{2} (26)$

$3X-2 =13
\\
3X = 15
\\
X =5$

Therefore

$GM = 3(5) -2 = 13$

And FM is 13 too .

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3 years ago

Which solution method would be most efficient to solve the equation (x+3)^2-4=0?

Angelina_Jolie [31]

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B-Factoring

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2 years ago

A rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1

Korvikt [17]

Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find : transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

4 0

3 years ago

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