The measure of central angle XYZ is 1.25 radians.

180 product of a prime factor using indices

zvonat [6]

Start with 180.
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>

<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>

<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>

<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>

<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>

<span>Once you end up with a quotient of "1" you're done. </span>

<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>

5 0

3 years ago

Read 2 more answers

Please help me guyys

algol13

Answer:

the answer is A! goodluck with hw thk

4 0

2 years ago

How do you solve 2sin^2θ = 1 ?

Murrr4er [49]

The value of θ is 45 degrees as sin45° is equal to 1/√2.

<h3>What is trigonometric equation?</h3>

Trigonometric equation refers to an equation consists of a number of trigonometric ratios whose angle is unknown.

<h3>What is the ranges of trigonometric angle?</h3>

Generally trigonometric angles are represented by θ. while solving equations we get the values of θ ranges from 0 degree to 90 degrees.

Given, 2sin²θ = 1

solving the equation

sin²θ = 1/2

sinθ = 1/√2

we know sin45° has a specific value 1/√2

sinθ = sin 45°

now, θ = 45°

hence, the solution is θ = 45 degrees.

to know more about trigonometric ratios and values visit

the link: brainly.com/question/29755110

#SPJ4

5 0

1 year ago

The cube has a volume of 27 in3. Find the volume of a scaled image with a scale factor of 2. in3 Cube with a side length of thre

WINSTONCH [101]

The new size of cube's side is 6. V new cube = 6*6*6 = 216

7 0

4 years ago

PRECAL:<br> Having trouble on this review, need some help.

ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

$\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}$

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

$\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}$

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

$\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}$

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

$\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}$

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

$\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}$

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

$\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}$

8. Divide through by x² again:

$\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}$

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

$\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}$

10. Factorize the numerator and simplify:

$\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2$

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

$\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}$

6 0

2 years ago