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torisob [31]
3 years ago
11

the rectangle below is shown with dimensions in inches the rectangle is tiled using squares that are 1/8 in each edge what is th

e area of the rectangle in square inches?
Mathematics
1 answer:
velikii [3]3 years ago
6 0

Answer:

Sorry i need points to ask a question sorry dont report pos

Step-by-step explanation:

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You make photo props by cutting rectangles out
KATRIN_1 [288]

Answer:

I can only read the top 2 dimensions:

The one on the left is 32 ft^2, the one on the right is 31 ft^2

Step-by-step explanation:

left top: (8x5)-(2x4) = 32

right top: (8x4)-(1x1) = 31

left low: (10x?)-(4x7) = ?

right low: (8x?)-(2x6) = ?

8 0
3 years ago
Solve: x + 1 < 4 and 3 - x > 5
Reika [66]

Answer:

okokok

Step-by-step explanation:

6 0
3 years ago
May you help me please I am stressed and struggling
alexandr402 [8]

Answer:

The answer would be 2x+4+3x+4, 5x+8, and 2(2x+4) +x

Step-by-step explanation:

Step 1. Simplify 2x+4+3x+4 into 5x (combine the x's) +8 (combine the 2 4's)
Step 2. Simplify 2(2x+4)= x into 4x + 8 + x (<-- distribute the 2 to the 2x and 4) and 4x+x would equal 5x meaning 5x + 8.
I hope that this helps! :)

3 0
2 years ago
Question part points submissions used use newton's method with the specified initial approximation x1 to find x3, the third appr
serious [3.7K]

Set f(x)=2x^3-3x^2+2. Find the tangent line \ell_1(x) to f(x) at the point when x=x_1:

f'(x)=6x^2-6x\implies f'(x_1)=12 (slope of \ell_1)

\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9

Set x_2=-\dfrac9{12}, the root of \ell_1(x). The tangent line \ell_2(x) to f(x) at x=x_2 has slope and thus equation

f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}

which has its root at x_3=-\dfrac{151}{224}\approx-0.6741.

(The actual value of this root is about -0.6777)

5 0
3 years ago
Please please help <br><br> Find the measure of the missing angle.<br> 63°
Ilya [14]

Answer:

27 degrees

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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