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1 year ago

Which of the following is the correct factorization of the polynomial below?

Mathematics

1 answer:

Alika [10]1 year ago

8 0

they easiest way to find the right answer is to look at the 2 first terms and checking if they equal 64x³

doing so:

48x³+...

64x³+...

9x³-48x³+...

= -39x³+...

-> the answer should be B, to make sure this is the right answer, you can just solve B

(4x+3)(16x²-12x+9)

= 64x³- 48x²+36x+48x²-36x+27

= 64x³+27

-> answer B is correct

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5x+4 HELP ME PLS it’s for a test I have right now

saveliy_v [14]

Answer:

Here is a graph of that equation.

Let me know if you need more help! :)

4 0

1 year ago

Read 2 more answers

The lsrl after an exponential transformation is log yhat=.4785 + 1.468x. What is the exponential form of the regression?

zheka24 [161]

Answer:

$10^{0.4785+1.468x}=y$

Step-by-step explanation:

So I'm assuming when you typed "log yhat=.4785 + 1.468x", you meant to write: $log(y)=0.4785+1.468x$ . And generally a logarithm can be written in the form $log_ba=c$ which can then be rewritten as $b^c=a$ , but since the log has no base, it's assumed to be 10. So in this case you have the equation:

$log_{10}y=0.4785+1.468x$ , which can then be written in exponential form as:

$10^{0.4785+1.468x}=y$

8 0

11 months ago

A dealer sold a car to a man and made

fenix001 [56]

Answer:

<u></u>

Step-by-step explanation:

Selling at a loss of 5% means it was sold for 95% of its purchase value.

If 120.175 equals to 95% then 100%(the purchase value) is 126.500.

120.175/95*100 = 126.500 (plenty of other ways to calculate this, but this is one of the easiest)

So now we change to the dealer's perspective, he sold the car for 126.500 and for him that meant a 15% profit. If the purchase value is 100%, then the sale value is 115% (purchase value+15% profit)

Calculate the purchase value:

126.500/115*100 = 110.000

The dealer bought the car for 110.000, sold it for 126.500, so he made a profit of 16.500 (126.500-110.000) which is a 15% profit (15% of 110.000 is 16.500).

5 0

2 years ago

-5 1/2 - 4 1/4 + 6 3/4

ruslelena [56]

Answer:

-3

Hope this helps :)

Have a great day !

5INGH

Step-by-step explanation:

$-5\frac{1}{2} - 4\frac{1}{4} +6 \frac{3}{4} = -5\frac{2}{4} - 4\frac{1}{4} +6 \frac{3}{4} = -9\frac{3}{4} +6\frac{3}{4}=-3$

5 0

2 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

2 years ago