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3 years ago

Please help!

Mathematics

1 answer:

Wittaler [7]3 years ago

7 0

Answer:

14 minutes

Step-by-step explanation:

10 times 14 = 140

15 times 14 = 210

200+210 = 410

550-140 = 410

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Let f(x) = 7x^2-5x+3 and g(x) = 2x^2+4x-6

Troyanec [42]

Answer:

$\large\boxed{A.\ f(x)+g(x)=9x^2-x-3}\\\boxed{B.\ f(x)-g(x)=5x^2-9x+9}\\\boxed{g(x)-f(x)=-5x^2+9x-9}$

Step-by-step explanation:

$f(x)=7x^2-5x+3,\ g(x)=2x^2+4x-6\\\\A:\\f(x)+g(x)=(7x^2-5x+3)+(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3+2x^2+4x-6\qquad\text{combine like terms}\\f(x)+g(x)=(7x^2+2x^2)+(-5x+4x)+(3-6)\\f(x)+g(x)=9x^2-x-3$

$B:\\f(x)-g(x)=(7x^2-5x+3)-(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3-2x^2-4x+6\qquad\text{combine like terms}\\f(x)-g(x)=(7x^2-2x^2)+(-5x-4x)+(3+6)\\f(x)+g(x)=5x^2-9x+9$

$C:\\g(x)-f(x)=(2x^2+4x-6)-(7x^2-5x+3)\\g(x)-f(x)=2x^2+4x-6-7x^2+5x-3\qquad\text{combine like terms}\\g(x)-f(x)=(2x^2-7x^2)+(4x+5x)+(-6-3)\\g(x)-f(x)=-5x^2+9x-9$

3 0

3 years ago

Find the distance between 13/3 and the square root of 16/9

Ainat [17]

Answer:

Root 16=4

Root 9=3

Therefore it becomes 4/3

Now distance between the two= 13/3-4/3 which is equal to 9/3 and 9 divided by 3 is 3.

Therefore the distance between the two on a number line is 3

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