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kolbaska11 [484]
3 years ago
14

Suppose sarah is trying to design a metal box (a rectangular prism), which will hold 200 cubic centimeters of liquid. the length

of the box must be 4 times as large as the width.
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
8 0
Volume = length * width * height = lwh (measured in cubic units)
so 2000 cm^3 = lwh
we know that l = 4w so sub that into the equation
2000 = (4w)*w*h
2000 = 4w^2*h

You didn't include the rest of the question so this is as far as i can go.
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Robinson's Doughnut Factory produces 18,000 doughnuts every day. They
Nadusha1986 [10]

Answer:

1,500 doughnuts

Step-by-step explanation:

Divide 18,000 by 12 to find how many will be loaded on each one:

18,000/12

= 1500

So, 1,500 doughnuts will be loaded on each truck

8 0
3 years ago
What is the trigonometric ratio for cos C?
Solnce55 [7]
You remember from SOH CAH TOA that
.. Cos = Adjacent/Hypotenuse
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8 0
2 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
I will mark that person brainliest if they answer correctly and must be well explained
Zolol [24]

it is 180

hope this helps

6 0
3 years ago
Read 2 more answers
Pls help <br><br> due at 11:59
evablogger [386]

Answer:

C. E.

Step-by-step explanation:

7 0
2 years ago
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