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slava [35]
2 years ago
14

What is the scale factor of AUVWto AXYZ?

Mathematics
2 answers:
Furkat [3]2 years ago
3 0

Step-by-step explanation:

Scale Factor of UVW to XYZ

= XY/UV = YZ/VW = XZ/UW

= 21/3 = 28/4 = 14/2

= 7. (B)

mixer [17]2 years ago
3 0
7, because UV(3)-XY(21), VW(4)-YZ(28), and WU(2)-ZX(14). To get from the first number in each set to the second number, multiply by 7. :)
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How to bypass school chromebook and download games?
charle [14.2K]

Answer:

chromebooks cant download games with or without a school filter

however, if you search "unblocked games" you might find some not blocked games

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help please i need this asap !!
Nookie1986 [14]

Answer:

slope is about 1.67

Step-by-step explanation:

2/1.2 ≈ 1.67

hope this helps :)

5 0
3 years ago
9Find -18 + (-67). Show your work.
polet [3.4K]

Answer:

-85

Explanation:

Given the mathematical expression:

-18+(-67)

First, we recall the product of signs.

+\times-=-

So, first, we open the bracket:

-18+(-67)=-18-67

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=-85

The result is -85.

3 0
1 year ago
A line includes the points (8,-2) and (-8,-4). What is the slope intercept
VladimirAG [237]

Answer:

The equation of the line would be y = 1/8x - 3

Step-by-step explanation:

to find the slope intercept form of the line, you must first find the slope. You can do this using the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (-2 - -4)/(8 - -8)

m = 2/16

m = 1/8

Then you can use this slope along with either point in point-slope form. Then solve for y.

y - y1 = m(x - x1)

y + 4 = 1/8(x + 8)

y + 4 = 1/8x + 1

y = 1/8x - 3

8 0
3 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
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