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Nuetrik [128]
2 years ago
8

Plsss help I need answers by today

Mathematics
1 answer:
pentagon [3]2 years ago
3 0

Answer:

E, A, F

Step-by-step explanation:

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A photographer made a 5-inch by 8-inch print of a photo. He then made a 10-inch by 16-inch print of the same photo. Which scale
GREYUIT [131]

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A.

Step-by-step explanation:

5x8 enlarged to 10x16

10 is 5x2 and 16 is 8x2

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Classify the following triangle. Check ALL that APPLY
weeeeeb [17]

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Naya [18.7K]

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y = 3/4 x + 25/4

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8 0
3 years ago
The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room
tangare [24]
I'm going to assume that the room is a rectangle.

The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle. 

You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
A = lw\\
600 = (x+5)(x+4)\\
600 =  x^{2} + 9x + 20


Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
600 = x^{2} + 9x + 20\\
x^{2} + 9x - 580 = 0\\
(x + 29)(x - 20) = 0\\
x + 29 = 0, \:\: x - 20 = 0\\
x = -29, x = 20

Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.

The dimensions of your room are 25ft (length) by 24ft (width).
8 0
3 years ago
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V-9/10v+6=11(hlp)this problem is confusing.
Valentin [98]
\frac{v-9}{10v+6}=11 \\\\ 11(10v+6)=v-9 \\\\ 110v+66=v-9 \\\\ 110v-v=-9-66 \\\\ 109v= -75 \\\\ \boxed{v=-\frac{75}{109}}
4 0
3 years ago
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