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alexandr1967 [171]
3 years ago
8

Solve for L: A = L ⋅ W ⋅ H (1 point)

Mathematics
2 answers:
butalik [34]3 years ago
7 0

Answer:

L= A/WH

Step-by-step explanation:

Off topic Btw it reminds me of area = length * width/height

Natasha_Volkova [10]3 years ago
4 0
L=a/w and that’s the answer
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What is the exact volume of the cylinder
GrogVix [38]
V=hpir^2
h=20
r=12

v=20pi12^2
v=20pi144
v=2880pi

2nd one is answer
3 0
3 years ago
Read 2 more answers
This is confusing :(
ValentinkaMS [17]

Answer:

154

Step-by-step explanation:

The important thing to recognize about finding the area for these types of figures is to make it into shapes that are easier to calculate, such as rectangles.

1) There are different ways of creating rectangles from this figure; however I will choose to cut the shape into two rectangles:

Imagine if the "L" was cut down from the top part of the "L" to the bottom, as to create to rectangles.

There will be a tall vertical rectangle and a small horizontal rectangle.

The tall rectangle will have a length of 17 and a width of 6

The smaller rectangle will have a length of 4 and a width of 13

2) Calculate the areas of the two rectangles and add them together:

Tall: 17 * 6 = 102

Small: 4 * 13 = 52

Total = 102 + 52 = 154

6 0
3 years ago
What is y-4=2(x+3) in standard form
Archy [21]

Answer:

2x - y = -10

Step-by-step explanation:

Standard form: x + y = n

y - 4 = 2x + 6

y = 2x + 6 + 4

y = 2x + 10

2x - y = -10

5 0
3 years ago
Find the slope of a line perpendicular to y= -6x+4
saveliy_v [14]

Answer:

<h2>-6</h2>

Step-by-step explanation:

y= -6x+4 is in y=mx +b

y= -6x+4 \\Slope =m = -6

5 0
3 years ago
A Bernoulli differential equation is one of the form
attashe74 [19]

y'-\dfrac5xy=\dfrac{y^5}{x^9}

Divide through both sides by y^5:

y^{-5}y'-\dfrac5xy^{-4}=x^{-9}

Now let z=y^{-4}, so that z'=-4y^{-5}y'. Then

-\dfrac{z'}4-\dfrac5xz=x^{-9}

z'+\dfrac{20}xz=-4x^{-9}

Multiply both sides by x^{20}:

x^{20}z'+20x^{19}z=\left(x^{20}z\right)'=-4x^{11}

Integrate both sides to get

x^{20}z=-\dfrac{x^{12}}3+C\implies z=-\dfrac1{3x^8}+\dfrac C{x^{20}}

Solve for y:

y=\left(\dfrac C{x^{20}}-\dfrac1{3x^8}\right)^{-1/4}

Given that y(1)=1, we have

1=\left(C-\dfrac13\right)^{-1/4}\implies C=\dfrac43

so the particular solution is

y=\left(\dfrac4{3x^{20}}-\dfrac1{3x^8}\right)^{-1/4}

which we can rewrite as

\boxed{y=\dfrac{3^{1/4}}{x^5\left(4-x^{12}\right)^{1/4}}}

4 0
3 years ago
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