All categories

3 years ago

Three cubes of each side 4cm are joining end to end. Find the surfacing area of resulting cuboid.

1 answer:

3 0

You would have to count the faces of cubes after you’ve joined them and then add all the surface areas up of each individual cube . I drew a pic to help you

You might be interested in

Find x Round your answer to the nearest integer. A. 7 B. 8 C. 6 D. 5

Ivenika [448]

Answer:

Explanations

in this case the hypotenuse is 10,x is the adjacent side of the angle 37

therefore you have to use cos

cos37=x/10

x=7.9

I hope this helps

3 0

3 years ago

Rewrite the equation in Ax+By=C form. Use integers A, B, and C. y+5=5(x-3)

Kaylis [27]

$y+5=5(x-3)\\ y+5=5x-15\\ 5x-y=20$

4 0

4 years ago

Read 2 more answers

Find the constant of variation for the relationship shown in the following table:

Lapatulllka [165]

The answer is y=4x because 1x4=4 2x4=8......

7 0

3 years ago

NEED NOW!!! Find the value of each variable in the diagram given that a ∥ b.

wolverine [178]

Answer:

W = Z = 180- 130 = 50

y = 30/2 = 15

x = 180 - 50 - (2x15) = 100

Step-by-step explanation:

W = Z = 180- 130 = 50

y = 30/2 = 15

x = 180 - 50 - (2x15) = 100

5 0

3 years ago

In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random

kupik [55]

Answer:

a) $\alpha = 3, \beta = 2$

b) 0.0620

Step-by-step explanation:

We are given the following in the question:

Population mean, $\mu$ = 6

Variance, $\sigma^2$ = 12

a) Value of $\alpha, \beta$

We know that

$\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12$

Dividing the two equations, we get,

$\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3$

b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

We can write the probability density function as:

$f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0$

We have to evaluate:

$P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620$

0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

3 0

4 years ago