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murzikaleks [220]
3 years ago
13

Given the two similar triangles below, which proportion is true?

Mathematics
2 answers:
gtnhenbr [62]3 years ago
5 0

Answer:

B)   10/15 = 15/22.5

Step-by-step explanation:

For the two triangles, proportions that involve ratios of the corresponding sides are true.

Harlamova29_29 [7]3 years ago
3 0
The answer is C to that question.
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1 5/6 changes into improper fraction
SVETLANKA909090 [29]
1 \cfrac{5}{6} =\cfrac{6 \cdot1+5}{6} =\cfrac{6+5}{6} =\cfrac{11}{6}
3 0
4 years ago
Read 2 more answers
Tasha is five less than twice as old as his brother Nathan together their ages add up to 25 how old is Tasha and his brother Nat
Zolol [24]
2x-5=25
Add 5 to both sides
2x-5+5=25+5
2x=30
Now, divide both sides to find x
2x/2=30/2
x=15
Nathan is 15 years old
To find Tasha's age, subtract 25 and 15
Therefore, Tasha is 10 years old
Hope I helped!
6 0
3 years ago
-6(x-5)-4(6x+4) =-76
slava [35]

Answer: x=3

Step-by-step explanation:

6 0
4 years ago
The sides of a square all have a side length of y. Write a simplified area function in terms of y for a rectangle whose length i
gizmo_the_mogwai [7]

The simplified area function in terms of y for a rectangle is f(y)=2 y^{2}+4 y

<u>Solution:</u>

Given that length of each side of a square = y

Need to determine area of rectangle whose length is twice the length of the square and width is 2 units longer that the side length of square

Length of rectangle = twice of side length of square = 2 \times y = 2y

Width of rectangle = 2 + side length of square = 2 + y = y + 2

\text { Area of rectangle }=\text { length of rectangle } \times \text { width of rectangle.}

On substituting length and width in formula for area, we get

\text { Area of rectangle }=2 y \times (y+2)=2 y^{2}+4 y

Hence function f(y)=2 y^{2}+4 y is represents area of required rectangle.

3 0
3 years ago
JK=4x+7 and KL=3x+14 find JL
Lena [83]

Answer:

Step-by-step explanation:

4x + 7 = 3x + 14

x + 7 = 14

x = 7

4(7) + 7 + 3(7) + 14

28 + 7 + 21 + 14

35 + 35 = 70

5 0
3 years ago
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