Please can you help me with this problems. Thanks

Someone help me pls

butalik [34]

the answer is A

5^-8 × 5^4= 5^-8+4

= 5^-4

= 1/5^4

8 0

3 years ago

What is the pattern in the values as the exponents increase?

Troyanec [42]

Divide the previous value by 3

6 0

3 years ago

Read 2 more answers

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

Solve y^2 4y - 32 = 0 using the zero product property.a. y=3b. y = -8, 4c. y = 8, -4d. y = -8, 6

Lady_Fox [76]

Y^2 + 4y - 32 = 0
y^2 - 4y + 8y - 32 = 0
y(y - 4) + 8(y - 4) = 0
(y + 8)(y - 4) = 0
y + 8 = 0 or y - 4 = 0
y = -8 or y = 4.

4 0

3 years ago

Which trigonometric function has a range that does not include 0.4?

Montano1993 [528]

<h2>Hello!</h2>

The answer is: $y=cscx$

Domain and range of trigonometric functions are already calculated, so let's discard one by one in order to find the correct answer.

The range is where the function can exist in the vertical axis when we assign values to the variable.

First:

$y=cosx$ : Incorrect, it does include 0.4 since the cosine range goes from -1 to 1 (-1 ≤ y ≤ 1)

Second:

$y=cotx$ : Incorrect, it also does include 0.4 since the cotangent range goes from is all the real numbers.

Third:

$y=cscx$ : Correct, the cosecant function is all the real numbers without the numbers included between -1 and 1 (y≤-1 or y≥1).

Fourth:

$y=sinx$ : Incorrect, the sine function range is equal to the cosine function range (-1 ≤ y ≤ 1).

I attached a pic of the csc function graphic where you can verify the answer!

Have a nice day!

8 0

4 years ago