Square RSTU is inscribed in circle P. Given the coordinates for the vertices of the square, find the equation of circle P.

Please help, it says NOT the correct way.

ololo11 [35]

Answer:

The first one or the 2nd one, since it says NOT worth half, I would say the first one is the answer.

Step-by-step explanation:

5 0

2 years ago

What is the y- intercept of the equation Y=2x+7?

expeople1 [14]

Answer:

y-intercept = 7

Step-by-step explanation:

The y-intercept is also the starting point

8 0

3 years ago

Which applies the power of a product rule to simplify (5 t) cubed?

Elena L [17]

Answer:

(5 t ) cubed = 5 cubed . t cubed = 125 t cubed applies the power of a product rule to simplify (5 t) cubed ⇒ 3rd answer

Step-by-step explanation:

Let us revise some rules of exponents

$a^{m}$ × $a^{n}$ = $a^{m+n}$

$a^{m}$ ×÷ $a^{n}$ = $a^{m-n}$

$(a^{m})^{n}$ = $a^{m.n}$

$(a.b)^{m}$ = $a^{m}$ . $b^{m}$

To simplify $(5t)^{3}$

∵ 5t means 5 × t

∵ Both of them are cubed

- Use the 4th rule above

∴ $(5t)^{3}$ = $(5)^{3}.(t)^{3}$

∵ (5)³ = 5 × 5 × 5 = 125

∴ $(5t)^{3}$ = $(5)^{3}.(t)^{3}$ = 125 t³

(5 t ) cubed = 5 cubed . t cubed = 125 t cubed applies the power of a product rule to simplify (5 t) cubed

3 0

3 years ago

Read 2 more answers

If you need 3/4 cups of beans for 7 servings. How many do you need if you want to make 35 servings

alukav5142 [94]

Answer:

3 3/4 cups

Step-by-step explanation:

We need 3/4 cups of beans for 7 servings, and we need to figure out how many cups we need for <em>35 servings</em>. You'll notice that 7 times 5 is 35. We can multiply 3/4 by 5 to find the answer to our problem. This thought process looks somewhat like this:

$\frac{\frac{3}{4} }{7} = \frac{x}{35}$

Notice that we can multiply 5 by 3/4 to find the value of x.

So, 3/4 times 5 is equal to 15/4, or 3 3/4. So, you need 3 3/4 cups of beans for 35 servings.

Hopefully this was helpful! If you have any more questions, let me know.

6 0

3 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

4 0

3 years ago

Read 2 more answers