Solve w^3=10 where w is a real number. Simplify your answer as much as possible.

I need help with this one

antiseptic1488 [7]

Answer:

-12

Step-by-step explanation:

3 0

3 years ago

HELP ASAP!!!<br><br><br> What system of inequalities is shown in the graph

storchak [24]

They need to be number C plus 3-2:7

4 0

3 years ago

Melody paid off a loan of $120,000 in 10 years. If she made a payment of $1,225 each month, then what was the simple interest ra

sdas [7]

I think it’s .01% because 1225/120000=.010208

7 0

3 years ago

Viviana had 4 and 1/5 pounds of sugar. She used 1/3 of the sugar she had. How much sugar did she use? Select all that apply

hram777 [196]

Answer:

11.339809 c

To convert a pound measurement to a cup measurement, multiply the sugar by the conversion ratio.

Since one pound of sugar is equal to 2.267962 cups, you can use this simple formula to convert:

cups = pounds × 2.267962

The sugar in cups is equal to the pounds multiplied by 2.267962.

For example, here's how to convert 5 pounds to cups using the formula above.

5 lb = (5 × 2.267962) = 11.339809 c

While experts usually suggest measuring dry ingredients by weight since it's more accurate, some recipes call for ingredients by volume and many of us don't have a scale when we need one. Because of the density of different types of sugar varies, it may not be obvious how to convert between a weight and volume measurements.

This table shows the approximate volume measurement for various weights of sugar, by type to help with the conversion.

Sugar Weight to Volume Conversion Table

Pound measurements and equivalent cups measurements for various types of sugar.

Should I Measure Sugar by Weight or Volume?

Many experts are adamant that dry ingredients like sugar should be measured by weight instead of volume, especially when used for baking.

Step-by-step explanation:

To convert a pound measurement to a cup measurement, multiply the sugar by the conversion ratio.

Since one pound of sugar is equal to 2.267962 cups, you can use this simple formula to convert:

cups = pounds × 2.267962

The sugar in cups is equal to the pounds multiplied by 2.267962.

For example, here's how to convert 5 pounds to cups using the formula above.

5 lb = (5 × 2.267962) = 11.339809 c

While experts usually suggest measuring dry ingredients by weight since it's more accurate, some recipes call for ingredients by volume and many of us don't have a scale when we need one. Because of the density of different types of sugar varies, it may not be obvious how to convert between a weight and volume measurements.

This table shows the approximate volume measurement for various weights of sugar, by type to help with the conversion.

Sugar Weight to Volume Conversion Table

Pound measurements and equivalent cups measurements for various types of sugar.

Should I Measure Sugar by Weight or Volume?

Many experts are adamant that dry ingredients like sugar should be measured by weight instead of volume, especially when used for baking.

<h2>Welcome for the answer. This answer is only for</h2><h2>good intentions please only report if necessary. Have a great rest of your day! </h2>

6 0

3 years ago

Find the moment of inertia about the y-axis of the thin semicirular region of constant density

arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

$\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA$

x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

$\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_x = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about the y-axis will be

$\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about O will be

$\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4$

More about the rotational inertia link is given below.

brainly.com/question/22513079

#SPJ4

3 0

2 years ago

Solve w^3=10 where w is a real number. Simplify your answer as much as possible.​

Solve w^3=10 where w is a real number. Simplify your answer as much as possible.