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Contact [7]
4 years ago
13

The total surface area of a cube is 294 cm ² work out the volume of the cube

Mathematics
2 answers:
mihalych1998 [28]4 years ago
6 0

\text{ Total Surface Area of cube, with Side $a$ is } 6a^2 \\ \text{ (six times the area of each face)} \\ 6a^2=294 \ m^2 \\ a^2= \frac{294}{6} \\ a^2=\frac{49\times6}{6} \\ a^2=49 \\ a=\sqrt{49} \\ a=7 \ m \\ \text{Volume of cube with Side $a$ is } a^3 \\ V= 7^3=49\times7 \\ \text{volume of the cube is } 343 \ m^3

melamori03 [73]4 years ago
3 0

At = 6×L² = 294 cm²

L² = 294cm²/6

L² = 49cm²

L = 7 cm

V = L³

= (7cm)³

= 343 cm³

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Better hearing and better sound censory

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3 years ago
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Given the function h(x) = 3(5)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.
otez555 [7]

We can write the function in terms of y rather than h(x) so that:

y = 3 (5)^x

 

A. The rate of change is simply calculated as:

r = (y2 – y1) / (x2 – x1) where r stands for rate

 

Section A:

rA = [3 (5)^1 – 3 (5)^0] / (1 – 0)

rA = 12

 

Section B:

rB = [3 (5)^3 – 3 (5)^2] / (3 – 2)

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B. We take the ratio of rB / rA:

rB/rA = 300 / 12

rB/rA = 25

 

So we see that the rate of change of section B is 25 times greater than A

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=6%5Csqrt%7B3%7D%2B2%281-%5Csqrt%7B27%7D%29%3D2" id="TexFormula1" title="6\sqrt{3}+2(1-\sqrt{27
skelet666 [1.2K]

Answer:

This is true.

Step-by-step explanation:

Step 1: see that √27 is the same as √3*3*3 and thus 3√3

Step 2: remove parenthesis so that 2(1-3√3) becomes 2 - 6√3

Step 3: observe that the equation now becomes 6√3 + 2 - 6√3 = 2

if you simplify any further, you are left with 0=0, which is true.

Comment if there is any step that requires more details...

3 0
3 years ago
Read 2 more answers
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

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Veseljchak [2.6K]
For lines to be perpendicular they would need to have opposite slopes, so for a slope of -3/7 the slope of a perpendicular line would be 7/3
7 0
3 years ago
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