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alekssr [168]
3 years ago
10

Giving Brainliest! No bots or using for points or I’ll report you. This is really needed!

Mathematics
2 answers:
AlexFokin [52]3 years ago
5 0

Answer:

1.) L=5 W=4 H=1. 2.)L=5 W=4 H=2. 3.)L=5 W=4 H=5

Step-by-step explanation:

you look at the sides and the front and count the cubic units

monitta3 years ago
5 0

Answer:

1.Length is 5 for all prisms, width is 4 for all prisms.For prism 1, the height is 1. for prism 2 is 2. For prism 3, the height is 5.

2.The volume for prism 1 is 20. The volume for prism 2 is 40. The volume for prism 3 is 100

3.The surface area for prism 1 is 18. The surface area For prism 2 is 36. The surface area for prism 3, is 90.

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Calculate the perimeter of the figure
larisa86 [58]

Perimeter is distance all around the side of the figure, so you would do:

8 + 7 + 19 + 3 + y + x

Plug in y and x that you solved before: 8 + 7 + 19 + 3 + 11 + 4

Add them!: 52


The perimeter is 52 meters.

:)

5 0
4 years ago
8-6x>-18 solve for x
EleoNora [17]

Answer:

x=5

Step-by-step explanation:

Well all i did was guess a random number from 1-10 so I could see what numbers make the equation bigger than -18 and the fist thing I guessed was 5 so this is what I did

8-6*5=-22

And them I said let me see if any multiples equal -18 so I went to for and did this

8-6*4=-16

That's when I guessed the answer if i looked at the sign correctly!

8 0
3 years ago
What is the equation of a circle with its center at (10,−4) and a radius of 2?
Dmitrij [34]
Equation of circle
(x-h)^2 + (y-k)^2 = r^2
=> (x-10)^2 + (y+4)^2 = 4 is the answer
7 0
3 years ago
Can anyone help i will give extra points
nataly862011 [7]

For question 3: 12 boys bc since 60 percent are girls, 40 percent are boys, the proportion 40/100=X/30 can be solved, 1200=100X, leaving you with X=12 boys

5 0
3 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
NNADVOKAT [17]

y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

4 0
3 years ago
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