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Lynna [10]
3 years ago
5

PLEASE HELP < I NEED HELP, TRYING TO MAKE HONOR ROLL

Mathematics
1 answer:
Stels [109]3 years ago
6 0

Answer: w = 4\frac{1}{3}

Step-by-step explanation:

Let the unknown number be w , then

the product of w and 3 = 3w

Nine less than the product of 3 and the number = 3w - 9

completing the equation , we have

3w - 9 = 4

add 9 to both sides

3w = 13

w = 13/3

w = 4\frac{1}{3}

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The length of a rectangle is 2 in longer than its width. if the perimeter of the rectangle is 44 in , find its area.
valkas [14]
P = 2(L + W)
P = 44
L = W + 2

44 = 2(W + 2 + W)
44 = 2(2W + 2)
44 = 4W + 4
44 - 4 = 4W
40 = 4W
40/4 = W
10 = W

L = W + 2
L = 10 + 2
L = 12

A = L * W
L = 12
W = 10

A = 12 * 10
A = 120 square inches <===
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3 years ago
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Question in pic. please help!
bonufazy [111]

Answer: 1, 3, and 4 can form triangles. The rest can not.

Step-by-step explanation:

on triangles 1, 3, and 4, the 2 smaller sides added together are greater than the longer side.

4 0
2 years ago
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What is the equation in point-slope form of a line that passes through the points (−4, −1) and (5, 7) ?
Sergio039 [100]

The point-slope form:

y-y_1=m(x-x_1)

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-4, -1) and (5, 7). Substitute:

m=\dfrac{7-(-1)}{5-(-4)}=\dfrac{7+1}{5+4}=\dfrac{8}{9}\\\\\boxed{y-7=\dfrac{8}{9}(x-5)}


6 0
2 years ago
Find the equation of a line parallel to line y=6 that passes through the point (-2,-1)
ohaa [14]

Answer:

y=-1

Step-by-step explanation:

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4 0
2 years ago
A solid lies between planes perpendicular to the​ x-axis at x = -2 and x = 2. The​ cross-sections perpendicular to the​ x-axis b
icang [17]

Answer:

V = 128π/3 vu

Step-by-step explanation:

we have that: f(x)₁ = √(4 - x²);  f(x)₂ = -√(4 - x²)

knowing that the volume of a solid is  V=πR²h, where R² (f(x)₁-f(x)₂) and h=dx, then

dV=π(√(4 - x²)+√(4 - x²))²dx;  =π(2√(4 - x²))²dx ⇒

dV= 4π(4-x²)dx , Integrating in both sides

∫dv=4π∫(4-x²)dx , we take ∫(4-x²)dx and we solve

4∫dx-∫x²dx = 4x-(x³/3) evaluated -2≤x≤2 or too 2 (0≤x≤2) , also

∫dv=8π∫(4-x²)dx evaluated 0≤x≤2

V=8π(4x-(x³/3)) = 8π(4.2-(2³/3)) = 8π(8-(8/3)) =(8π/3)(24-8) ⇒

V = 128π/3 vu

8 0
3 years ago
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