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3 years ago

Draw an angle measuring 700 degrees in standard position.

Mathematics

1 answer:

3241004551 [841]3 years ago

7 0

A screenshot of the angle is attached.

700 is almost 2 complete revolutions around the graph. 700-360=340; this means it is the same as going 20° backwards from the starting point.

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Sylvia earns $7 per hour at her after school job. After working one week she recieved a paycheck for $91. Write and solve an equ

Mashutka [201]

Amount of hours worked = h
7h = 91
h = 13 hours worked that week
h (is greater to or equal to) 15
7 x 15 would equal 105 which means the most she can make in a week is $105

3 0

3 years ago

Gimme the answer because I have no idea how to do this it makes no sense to me

kati45 [8]

Answer:

C, D, B

Step-by-step explanation:

the mode is if a number is repeated more than one time

EX. 15, 23, 15, 23, 15

C because 15 is repeated 3 times and 19 is too.

D because 42 is repeated 2 times and so is 18.

B because 87 is repeated 2 times and 32 is too.

5 0

3 years ago

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti

Vikki [24]

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$ ..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$

Thus the solution becomes

$\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c$ $y=x^{3}+cx^{2$

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have $6=8+4c$

Thus solving for c we get c = -1/2

Thus particular solution becomes

$y=x^{3}-\frac{1}{2}x^{2}$

5 0

3 years ago

1 point

Ber [7]

15
Explanation:
you just divide 90 by 6

6 0

2 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Hence, the required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

3 0

3 years ago