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kakasveta [241]
2 years ago
11

What times what equals -152 and when added equals 12

Mathematics
1 answer:
sergeinik [125]2 years ago
6 0
You can set up an equation to figure this out.

-152+x=12

Add 152 to both sides and you get:

x=164
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Please write out the problem on paper
skad [1K]
Your answer will be n5
7 0
2 years ago
A figure has vertices at L(3,3), M(2,4), and N(−1,0). After a transformation, the image of the figure has vertices at L′(2,−5),
alexandr402 [8]

The transformation was a reflection over the line x = -0.5

And the image and preimage can be seen in the graph below.

<h3>How to identify the preimage and image?</h3>

First, the original vertices are:

L(3, 3)

M(2, 4)

N(-1, 0)

3 vertices means that our figure is a triangle.

Now, the transformed vertices are:

L'(2, -5)

M'(3, -4)

N'(-1, 1)

And the graph of both sets can be seen below.

Now we can try to identify the transformation.

Now, notice that all the points (L, M, N, L', N', M') are equidistant to the horizontal line x = -0.5

This is a clear implication that the transformation was a reflection over the line x = -0.5

If you want to learn more about transformations:

brainly.com/question/4289712

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3 0
1 year ago
Find the area and perimeter.
lara [203]

Answer:

Step-by-step explanation:

Perimeter of triangle =  4x - 3 + x + 9 + 8

                                   = <u>4x + x</u> -<u> 3 + 9 + 8</u>

                                  = 5x + 14

Area  = \frac{1}{2}*base *height

         = \dfrac{1}{2}*8*3x\\\\= 4 * 3x\\\\= 12x

8 0
2 years ago
10 binder; 60% off; 6% tax
laila [671]

Answer:

I'm going to assume that the binder is $10, and that the tax is applied after the 60% off, in which case the answer is $4.24

Step-by-step explanation:

10 * 10 = 100

6 * 10 = 60

10 - 6 = 4

6 / 25 = 0.24

4 + 0.24 = 4.24

Answer: $4.24

4 0
2 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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