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3 years ago

PLEASE ANSWERRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR

Mathematics

1 answer:

Lesechka [4]3 years ago

8 0

Answer:

So the 1st, 2nd, & 3rd graphs all show continuous data. (I'll attach an image to show which ones I'm talking about)

Explanation:

This is because with continuous data, all points need to connect and it needs to continue. The 4th graph (with just points) would be discrete.

I hope this helps!! :)

Sorry I took so long...

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A sphere and a cylinder have the same radius and height. The volume of the cylinder is 21m. what is the volume of the sphere.

Goshia [24]

Answer:

The volume of the sphere is 14m³

Step-by-step explanation:

Given

Volume of the cylinder = $21m^3$

Required

Volume of the sphere

Given that the volume of the cylinder is 21, the first step is to solve for the radius of the cylinder;

<em>Using the volume formula of a cylinder</em>

The formula goes thus

$V = \pi r^2h$

Substitute 21 for V; this gives

$21 = \pi r^2h$

Divide both sides by h

$\frac{21}{h} = \frac{\pi r^2h}{h}$

$\frac{21}{h} = \pi r^2$

The next step is to solve for the volume of the sphere using the following formula;

$V = \frac{4}{3}\pi r^3$

Divide both sides by r

$\frac{V}{r} = \frac{4}{3r}\pi r^3$

Expand Expression

$\frac{V}{r} = \frac{4}{3}\pi r^2$

Substitute $\frac{21}{h} = \pi r^2$

$\frac{V}{r} = \frac{4}{3} * \frac{21}{h}$

$\frac{V}{r} = \frac{84}{3h}$

$\frac{V}{r} = \frac{28}{h}$

Multiply both sided by r

$r * \frac{V}{r} = \frac{28}{h} * r$

$V = \frac{28r}{h}$ ------ equation 1

From the question, we were given that the height of the cylinder and the sphere have equal value;

This implies that the height of the cylinder equals the diameter of the sphere. In other words

$h = D$ , where D represents diameter of the sphere

Recall that $D = 2r$

So, $h = D = 2r$

$h = 2r$

Substitute 2r for h in equation 1

$V = \frac{28r}{2r}$

$V = \frac{28}{2}$

$V = 14$

Hence, the volume of the sphere is 14m³

4 0

3 years ago

Read 2 more answers

Internet providers: In a survey of 935 homeowners with high-speed Internet, the average monthly cost of a high-speed Internet pl

BlackZzzverrR [31]

Answer:

68% of plans cost between $62.16 and $86.52.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean: 74.34

Standard deviation: 12.18

Bell-shaped is the same as normally distributed.

Estimate the number of plans that cost between $62.16 and $86.52.

62.16 is one standard deviation below the mean.

86.52 is one standard deviation above the mean.

By the Empirical rule, 68% of the measures are within 1 standard deviation of the mean.

68% of plans cost between $62.16 and $86.52.

7 0

3 years ago

I need help please.

Oduvanchick [21]

Answer:13/34

Step-by-step explanation:

6 0

3 years ago

Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

$\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}$

Consider a vector field $F=x\^i+y\^j+z\^k$

Then the divergence of the vector F is,

div F = $\nabla.F$ = $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)$

and the curl of the vector F is,

curl F = $\nabla\times F$ = $\^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})$

<h3>Calculation:</h3>

The given vector fields are:

$F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$

1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

⇒ $\nabla.(aF1+bF2)$

⇒ $\nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))$

⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

From (1) and (2),

$\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

2) Verifying the identity: $\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Consider L.H.S

⇒ $\nabla\times(aF1+bF2)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the cross product,

$\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})$ ...(3)

Consider R.H.S,

⇒ $a\nabla\times F1+b\nabla\times F2$

So,

$a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))$

⇒ $\^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})$

$a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))$

⇒ $\^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})$

Then,

$a\nabla\times F1+b\nabla\times F2$ =

...(4)

Thus, from (3) and (4),

$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

8 0

1 year ago

11.(02.03)

Marianna [84]

Answer:

step number 1 is incorrect.

Step-by-step explanation:

Here is the solution of this equation [2x-31-1=2].

2x-31-1=2

or 2x-31=2+1

or 2x=2+1+31

or 2x= 34

or x=34/2

or x= 17

the answer of this equation is X=17.

8 0

3 years ago