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3 years ago

What are the negative and positive coterminal angles of 240 degrees?

Mathematics

2 answers:

algol133 years ago

7 0

-120 negative and 600 positive.

Norma-Jean [14]3 years ago

3 0

Answer:

Add 360° to find a positive coterminal angle. Subtract 360° to find a negative coterminal angle. Angles that measure 240° and –480° are coterminal with a –120° angle.

Step-by-step explanation:

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H e l p m e ! Please, man, mth is unforgivably hard and I can't fail!

Scorpion4ik [409]

Answer:

Step-by-step explanation:

Let's solve for x.

−3x+2y=−10

Step 1: Add -2y to both sides.

−3x+2y+−2y=−10+−2y

−3x=−2y−10

Step 2: Divide both sides by -3.

−3x /−3 = −2y−10 −3

x= 2 /3 y+ 10 /3

Answer:

x= 2 /3 y+ 10 /3

6 0

4 years ago

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

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

3 0

3 years ago

PLEASE HURRY WILL GIVE BRAINLIEST A truck can be rented from Company A for $90 a day plus $0.30 per mile. Company B charges $70

arlik [135]

After 40 miles the rental cost for both is the same

4 0

3 years ago

Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The m

svp [43]

Answer:

As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.

Step-by-step explanation:

As we suspect the variability of concentration F - test is applied.

n1=10 s1=4.7

n2=16 s2=5.8

And α = 0.05.

The null and alternate hypothesis are

H0: σ₁²=σ₂² Ha: σ₁²≠σ₂²

The null hypothesis is the variability in concentration does not differ for the two companies.

against the claim

the variability in concentration may differ for the two companies

The critical region F∝(υ1,υ2) = F(0.025)9,15= 3.12

and 1/F∝(υ1,υ2) = 1/3.77= 0.26533

where υ1= n1-1= 10-1= 9 and υ2= n2-1= 16-1= 15

Test Statistic

F = s₁²/s₂²

F= 4.7²/5.8²=0.6566

Conclusion :

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3 years ago

How do u change inches into feet

Mumz [18]

Answer:

12 INCHES = 1FOOT

Step-by-step explanation:

8 0

3 years ago

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