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wolverine [178]
2 years ago
8

What equation matches the known dimensions of the prism below? Prism with a volume of thirty six cubic inches; ? in., 4 in., 3 i

n. 4 × 3 × ? = 36 36 × ? × 4 = 3 4 × 3 × 36 = ? 36 × 3 × ? = 4
Mathematics
1 answer:
tatiyna2 years ago
7 0

Answer:

4 x 3 x ? = 36

Step-by-step explanation:

36 is quotient and we have 3 other numbers one we don't know so we simply do 4 x 3 x ? or the other way around.

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Need help finding out what 3x=y and x-2y = 10
sweet-ann [11.9K]

Answer:

x=-2, y=-6

Step-by-step explanation:

Simple substitution.

Since y is isolated in one of the equations already, you can plug it into the second equation.

x - 2(3x) = 10

x - 6x = 10

-5x = 10

x = -2

Sub. x = -2 into 3x = y:

3(-2) = y

y = -6

4 0
3 years ago
The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
ankoles [38]

Answer:

     2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

  T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

  D_x=T_x\sqrt{T_x}

The mean distance of planet Y can be found using the given relation ...

  T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x

The mean distance of planet Y is increased from that of planet X by the factor ...

  2√2

8 0
2 years ago
What is the solution to the equation 3/<br> x+4+3/2x+8 = 0?
mixas84 [53]

Answer:

X= -4

Step-by-step explanation:

^ you have to isolate the radical than raise each side to the power of its index which gets you

x= -4

5 0
3 years ago
A linear relation is given by the function f (x) = −7x +18 . If (x, 46) is on the function, the value of x is
prohojiy [21]

Answer:

x=-4

Step-by-step explanation:

46 = -7x + 18

subtract 18 from each side to get:

28 = -7x

divide each side by -7 to get:

x = -4

4 0
3 years ago
A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach
nlexa [21]

Answer:

A score of 150.25 is necessary to reach the 75th percentile.

Step-by-step explanation:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that \mu = 130, \sigma = 30

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 130}{30}

X - 130 = 0.675*30

X = 150.25

A score of 150.25 is necessary to reach the 75th percentile.

7 0
3 years ago
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