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kotegsom [21]
3 years ago
12

Solve for the variable

Mathematics
1 answer:
krek1111 [17]3 years ago
5 0

Answer:

G = 2/3

Step-by-step explanation:

-13G + 10 = 2(-5G + 4)  Distribute the 2 to (-5G + 4)

-13G + 10 = -10G + 8  

+13G           +13G         Add 13G to both sides

10 = 3G + 8

-8           -8                  Subtract 8 from both sides

2 = 3G                          Divide both sides by 3

2/3 = G

If this answer is correct, please make me Brainliest!

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This can be solved by finding their point of intersection on the graph, which is at point (1,3).
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3 years ago
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Two polygons are similar. The perimeter of the larger polygon is 120 yards and the ratio of the corresponding side lengths is 1/
Kay [80]

Answer:

20 yards

Step-by-step explanation:

Given that:

Two Given polygons are similar :

Ratio of corresponding sides =. 1/6

Perimeter of larger polygon = 120 yards

Perimeter of smaller = p

Since they are similar, and yhe ratio of their sides Given, we use the relation :

(Smaller perimeter / larger perimeter) = (smaller side / larger side)

(p / 120) = (1 /6)

Cross multiply :

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p = 20 yards

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3 years ago
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Irina-Kira [14]

Answer:

the answer is 0 to 1 and 3 to 4

7 0
3 years ago
The sum of nine-tenths and g is two-thirds solve for g
Naya [18.7K]

Answer:

- 7/30

Step-by-step explanation:

9/10 + g = 2/3

So subtract 9/10 from both sides of the equation to get g = -7/30

Thank me later :D

4 0
3 years ago
Someone please help or i’ll be failing geometry this year
Vlad [161]

Step-by-step explanation:

I don't know what constructions you were taught.

a "similar" triangle is a triangle with exactly the same angles as the other triangle, but the lengths of all sides are stretched or shortened by the same scaling factor f.

by saying 1:2 she means the second triangle should have sides with twice the lengths of the first triangle (f=2).

and the extra challenge - same basic thing. she allows you to pick one of the two triangles as reference. and then you need to draw a third triangle (again with the same angles) with the side lengths extended by the scaling factor f of 4/3.

I would draw the triangles right on top of each other with the same starting corner (let's call it A) for all 3.

we would get the triangles ABC, AMN and AXY.

the points B and C would be then halfway on AM and AN.

and M and N would then a bit before X and Y on AX and AY.

the beauty is, you only need to construct 2 sides of every new triangle down to the new endpoints. the third side is automatically scaled correctly, and you only need to connect these new endpoints.

let's assume you draw a triangle (just very simple) ABC with all side lengths being 3. so, AB=3, AC=3, BC=3.

now you draw AMN by extending AB and AC to a side length of 6 (f=2) creating M and N, and you connect M and N.

and then you can create the third triangle AXY by extending AM and AN by a factor of 4/3 to side lengths of 8 (4/3 × 6 = 8) creating new end points X and Y. and you connect X and Y.

and that is it. all 3 triangles are similar (the same angles), and all sides of a triangle have the same length ratio to the sides of the other triangle(s).

4 0
3 years ago
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