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3 years ago

Find the measure of angle R to the nearest tenth

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

The angles add up to 180°
59+39=98
180-98=82°

Alex3 years ago

4 0

Answer:

Step-by-step explanation:

I am going to use the sin law to find the angle R. Applying sin law we have that

$\frac{sin(90)}{58} = \frac{sin(R)}{37}$ , then

$sin(R) = \frac{sin(90)*37}{58} = \frac{37}{58} = 0.6379$

$R = sin^{-1}(0.6379) = 39.63\textdegree = 39.6\textdegree$ .

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Cedarburg's zoo has two elephants. The male elephant weighs 3 3/4 tons and the female

gtnhenbr [62]

Answer:

3.16666667 tons

Step-by-step explanation:

hope this helps have a good evening

3 0

3 years ago

John receives two credit card offers in the mail. Credit Card A: 1% introductory APR, 19.9% APR for purchases after the introduc

musickatia [10]

Answer:

(D) "...because he will not have to pay any interest on his purchases during the introductory APR period."

Step-by-step explanation:

There may be many reasons for John to choose one over the other depending on his future and planned usage patterns. However, answer the (D) is the only correct one in terms of stating a reason that is actually supported by the information in the question. Specifically, there is 0% APR for card B.

What's wrong with the other choices:

Choice A: incorrect statement re APR after intro period (is not 1%); B: incorrect statement re cash advance (is not 0%); C: statement is not supported by anything in the information given (it is an assumption, and is unlikely).

5 0

3 years ago

Read 2 more answers

| 2p -1 | =45, what are the values of p?

Alex73 [517]

Answer:

p = 23 and p = -22

Step-by-step explanation:

| 2p -1 | =45

Absolute value equations have 2 solutions, one positive value and one negative

2p-1 = 45 and 2p-1 = -45

Add 1 to all sides

2p-1+1 = 45+1 and 2p-1+1 = -45+1

2p = 46 2p = -44

Divide by 2

2p/2 = 46/2 and 2p/2 = -44/2

p = 23 and p = -22

5 0

3 years ago

Read 2 more answers

school teacher and needs 22 pieces of wood, 3 over 8 ft long, for a class project. If he has a 9-ft board from which to cut the

tiny-mole [99]

Answer:

Yes, he will have enough 3 over 8 ft pieces for his class.

Step-by-step explanation:

Given:

Number of wood required = 22

Length of each wood, $l=\frac{3}{8}\textrm{ ft}$

Total length of the board, $L=9\textrm{ ft}$

Therefore, the number of woods that can be made using the given board is given as:

$\textrm{Number of woods made}=\frac{\textrm{Total length of board}}{\textrm{Length of each wood}}\\\textrm{Number of woods made}=\frac{L}{l}=\frac{9}{\frac{3}{8}}=9\times \frac{8}{3}=\frac{72}{3}=24$

So, he can make 24 woods of length $\frac{3}{8}\textrm{ ft}$ using the 9 ft board. But he has to make only 22 pieces.

Therefore, he has enough of the wood to make the required number of pieces.

3 0

3 years ago

Trouble finding arclength calc 2

kiruha [24]

Answer:

$S\approx1.1953$

Step-by-step explanation:

So we have the function:

$y=3-x^2$

And we want to find the arc-length from:

$0\leq x\leq \sqrt3/2$

By differentiating and substituting into the arc-length formula, we will acquire:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx$

To evaluate, we can use trigonometric substitution. First, notice that:

$\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx$

Let's let y=2x. So:

$y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx$

We also need to rewrite our bounds. So:

$y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0$

So, substitute. Our integral is now:

$\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Let's multiply both sides by 2. So, our length S is:

$\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy$

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

$y=a\tan(\theta)$

Substitute 1 for a. So:

$y=\tan(\theta)$

Differentiate:

$y=\sec^2(\theta)\, d\theta$

Of course, we also need to change our bounds. So:

$\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0$

Substitute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta$

The expression within the square root is equivalent to (Pythagorean Identity):

$\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta$

Simplify:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta$

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

$u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta$

And:

$dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)$

Integration by parts:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)$

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)$

Distribute:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)$

Now, let's make the single integral into two integrals. So:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Distribute the negative:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

$\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta$

Divide the second and third equation by 2. So: $\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)$

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

$\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))$

Therefore, our arc length will be equivalent to:

$\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Divide both sides by 2:

$\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}$

Evaluate:

$S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))$

Evaluate:

$S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))$

Simplify:

$S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}$

Use a calculator:

$S\approx1.1953$

And we're done!

7 0

3 years ago