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3 years ago

Solve for g in the proportion. 3/g = 24/16

Mathematics

2 answers:

Nata [24]3 years ago

5 0

Answer:

This is your answer ☺️☺️☺️

strojnjashka [21]3 years ago

3 0

Answer:

Step-by-step explanation:

Hello,

We can solve this by simply cross multiplying both sides of the equation. This will give the following:

16x3 = 24xg

48 = 24g

Now solve for g by dividing both sides by 24. This will give the following:

g = 2

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The two figures are similar. Find the ratios (red to blue) of the perimeters and of the areas. Write the ratios as fractions in

vesna_86 [32]

Answer:

see explanation

Step-by-step explanation:

The perimeters of the similar figures have the same ratio as the sides.

ratio of perimeters = $\frac{11}{6}$

ratio of areas = 11² : 6² , that is 121 : 36

ratio of areas = $\frac{121}{36}$

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3 years ago

What is slope between the points (-5,7) and (4,-8)?

WINSTONCH [101]

The slope can sometimes be called the gradient, and the equation for the gradient is (y2 - y1)/(x2 - x1). So therefore, you'd do: (-8 - 7)/(4 - -5) which is (-15)/9) which is -1 2/3 or -1.6 (recurring), which is your answer. I hope this helps! Let me know if I've confused you :)

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4 years ago

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lakkis [162]

Answer:

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Step-by-step explanation:

4 0

3 years ago

Is a + b rational or irrational

Crank

Answer and fdaStep-by-step explanation:::

The answer depends on the numbers that are inputted for a and b.

If the numbers that are inputted for both a and b are rational, then a + b would be rational.

If the numbers that are inputted for both a and b are irrational, then a + b would be irrational.

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3 years ago

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

yan [13]

Each curve completes one loop over the interval $0\le t\le2\pi$ . Find the intersections of the curves within this interval.

$6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi$

The region of interest has an area given by the double integral

$\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta$

equivalent to the single integral

$\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta$

which evaluates to $9\pi+72$ .

8 0

3 years ago