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never [62]
3 years ago
6

So umm I’m dumb and I need help with this problem

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

A = 17.5

Step-by-step explanation:

Step 1: Find the base

Base = 7 units

Step 2: Find the height

Height = 5 units

Step 3: Plug your 2 variables into the Area Formula

A = 1/2bh

A = 1/2(7)(5)

A = 1/2(35)

A = 35/2

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retail=10.82(100+72)/100≈$18.61 (to nearest cent)
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2 years ago
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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
Ede4ka [16]

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

5 0
3 years ago
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1c=3s
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Crazy boy [7]

Answer:

I believe its 24 minutes or 30 minutes

Step-by-step explanation:

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3 years ago
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ANSWER THE FOLLOWING:​
Bogdan [553]

Answer:

1. (a) 1/12

(b)35/36

(c) 31/36

2. Blue - 1/3

Red- 4/15

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