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3 years ago

Can someone help me with this please.

Mathematics

1 answer:

allsm [11]3 years ago

3 0

It’s x is greater than or equal to two. the > with a line under it.

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Vertical stretch of the graph f(x)= x² by a factor of 5<br><br>PLEASE HELP!!

andreev551 [17]

Answer:

f(x) = 5x^2

Step-by-step explanation:

A vertical stretch is in the y direction

y = Cf(x) C > 1 stretches it in the y-direction

f(x) = 5x^2

6 0

3 years ago

At the begging of the month a store had a balance of -554

makvit [3.9K]

What is the question? Do you mean "At the beginning of the month a store had the balance of $554"? If so, what is the question?

5 0

4 years ago

Evaluate integral _C x ds, where C is

borishaifa [10]

Answer:

a. $\mathbf{36 \sqrt{5}}$

b. $\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

Step-by-step explanation:

Evaluate integral _C x ds where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

$\int \limits _c \ x \ ds$

where;

x = t , y = t/2

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt}= \dfrac{1}{2}$

and t varies from 0 to 12.

we all know that:

$ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt$

∴

$\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt$

$= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt$

$= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0$

$= \dfrac{\sqrt{5}}{4}\times 144$

= $\mathbf{36 \sqrt{5}}$

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t ; y = 3t²

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt} = 6t$

$ds = \sqrt{1+36 \ t^2} \ dt$

Hence; the integral _C x ds is:

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

Let consider u to be equal to 1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

$tdt = \dfrac{du}{76}$

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

$\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}$

$= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}$

$\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}$

$\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

5 0

4 years ago

Find the nth term of the sequence 7, 17, 27, 37

dangina [55]

Answer:

Step-by-step explanation:

7 plus 10 = 17, 17 plus 10, 27 and it goes on. pls give brainliest

5 0

3 years ago

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What is the area of this circle ( express your answer by using pi)

Leona [35]

Answer: option C

Step-by-step explanation:

You need to use the formula for calculate the area of a circle:

$A_{(circle)}=\pi r^2$

Where the radius of the circle is "r".

You can see in the figure that the diameter of the circle is 6 meters. Then you need to find the radius of the circle with the formula:

$r=\frac{d}{2}$

Where "d" is the diameter of the circle.

Then, the radius is:

$r=\frac{6m}{2}=3m$

Finally, substitute the radius into the formula. Then, the area of the circle is:

$A_{circle}=\pi (3m)^2=9\pi m^2$

7 0

4 years ago

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