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Lena [83]
3 years ago
12

A square picture frame measures 36 inches on each side. The actual wood trim is 2 inches wide. The photograph in the frame is su

rrounded by a bronze mat that measures 5 inches. What is the maximum area of the photograph

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
7 0
(36-2-5)^{2}=\\(29)^{2}=\\841

841 inches^2.
OverLord2011 [107]3 years ago
7 0
We need to figure out how much of the total area of the picture frame is usable. To do this, we need to figure out what is taken up by the frame and the mat.

The frame is 36 inches on a side.
We know the frame is 2 inches. We know that the the mat is 5 inches.

This means that on each 36 inch side, 4 inches are taken up by the frame (2 each side) and 10 inches by the mat (again, 5 each side). So the total length left is
36 - 4 - 10 = 32 - 10 = 22.

The maximum area of the photograph is therefore 22 x 22 = 484 square inches.

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What is the answer for the picture that is linked below?
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Can y’all give me some awnsers and number to them?
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Answer:

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5 0
2 years ago
1.
butalik [34]

The answers to each of the given problems are;

1) Sum of two smallest integers  = 23

2) 3x + 6

3) 7.5 m/s²

<h3>How to find the sum of consecutive Integers?</h3>

1) We are told that sum of 4 consecutive integers is increased by 20 and equals 70. Thus, we have;

x + (x + 1) + (x + 2) + (x + 3) + 20 = 70

4x + 26 = 70

4x = 70 - 26

4x = 44

x = 44/4

x = 11

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2) Let the consecutive odd numbers be;

x, (x + 2) and (x + 4)

Sum of these consecutive odd numbers is;

x + x + 2 + x + 4 = 3x + 6

3) We are given the equation to find the acceleration as;

(v_final)² - (v_initial)² = 2ad

We are given;

v_final = 40 m/s

v_initial = 10 m/s

d = 100 m

Thus;

40² - 10² = 2a(100)

1500 = 200a

a = 1500/200

a = 7.5 m/s²

Read more about sum of integers at; brainly.com/question/17695139

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