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Darina [25.2K]
3 years ago
15

Find the lengths of g, h, and j. Round answers to the nearest tenth. (marking brainliest for correct)

Mathematics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

j=13, g=20.8, h=24

Step-by-step explanation:

The overall shape given and the shape within, are both right triangles. With right triangles, you are allowed to use the pythagorean theorem formula (a^{2} + b^{2} = c^{2}) in order to solve for some sides. In this case, that would be j and h. The five in the smaller triangle is represented by b and the 12 is the hypotenuse so it is represented by c. When you plug in those numbers in the pythagorean theorem formula, you will find the value of j to be 13. When looking at this, we see that 12 is the second greatest value in the right triangle values that we just found, so we know the the opposing angle for that one will be 60 degrees. The 5's opposing side is therefore 30 degrees. When subtracting 90 and 30, we get 60, so therefore you can use the 30 60 90 formula to find the sides of the bigger triangle. The 60 degrees represents g. This formula will be a\sqrt{3}. The a is 12 since it is the smallest value. So therefore, g is 12\sqrt{3}, which is 20.8. Now that we have this side, we can just use the pythagorean theorem formula to find the remaining side. Therefore, h is going to be 24

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Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge
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Answer:

Part a) The quadratic function is 4x^{2} +20x-39=0

Part b) The value of x is 1.5\ in

Part c) The photo and frame together are 7\ in wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0

Part b) What is the value of x?

Solve the quadratic equation 4x^{2} +20x-39=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem

we have

4x^{2} +20x-39=0

so

a=4\\b=20\\c=-39

substitute in the formula

x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}

x=\frac{-20(+/-)\sqrt{1,024}} {8}

x=\frac{-20(+/-)32} {8}

x=\frac{-20(+)32} {8}=1.5\ in  -----> the solution

x=\frac{-20(-)32} {8}=-6.5\ in

Part c) How wide are the photo and frame together?

(4+2x)=4+2(1.5)=7\ in

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katen-ka-za [31]

Answer: £54.60

Step-by-step explanation:

7.80 x 7 = £54.60

***If you found my answer helpful, please give me the brainliest. :) ***

6 0
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