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Aleks [24]
3 years ago
10

Two negative integers are 5 units apart on the number line, and their product is 126. What is the sum of the two integers?

Mathematics
2 answers:
Ilya [14]3 years ago
7 0

Answer:-23

Step-by-step explanation:

Hope it helps

Vladimir [108]3 years ago
4 0

Answer:

-23

Step-by-step explanation:

-23 is the answer

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Answer:

Step-by-step explanation:

B (0,0)    ;  A (5,8)

AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\=\sqrt{(5-0)^{2}+(8-0)^{2}}=\sqrt{5^{2}+8^{2}}

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3 years ago
Order 1.09,1.9,and 1.1 from least to greatest
zysi [14]
1.09, 1.1, 1.9 Hope I helped
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CAN SOMEONE PLEASE HELP ME WITH THESE PEOPLE​
vazorg [7]

Inverse of f(x)=x+2 is f^{-1}(x)=x-2 because

f(x)=x+2\Rightarrow x=f^{-1}(x)+2\Rightarrow\boxed{f^{-1}(x)=x-2}

Now f^{-1}(-5)=-5-2=\boxed{-7}.

Hope this helps.

4 0
3 years ago
Investigate the range of each parent function below over the interval o ranges from least to greatest.<br> Identity function, re
Y_Kistochka [10]

Answer:

Here we have the domain:

D = 0 < x < 1

And we want to find the range in that domain for:

1) y = f(x) = x

First, if the function is only increasing in the domain (like in this case) the minimum value in the range will match with the minimum in the domain (and the same for the maximums)

f(0) = 0 is the minimum in the range.

f(1) = 1 is the maximum in the range.

The range is:

0 < y < 1.

2) y = f(x) = 1/x.

In this case the function is strictly decreasing in the domain, then the minimum in the domain coincides with the maximum in the range, and the maximum in the domain coincides with the minimum in the range.

f(0) = 1/0 ---> ∞

f(1) = 1/1

Then the range is:

1 < x.

Notice that we do not have an upper bound.

3) y = f(x) = x^2

This function is strictly increasing, then:

f(0) = 0^2 = 0

f(1) = 1^2 = 1

the range is:

0 < y < 1

4) y = f(x) = x^3

This function is strictly increasing in the interval, then:

f(0) = 0^3 = 0

f(1) = 1^3 = 1

the range is:

0 < y < 1.

5) y = f(x) = √x

This function is well defined in the positive reals, and is strictly increasing in our domain, then:

f(0) = √0 = 0

f(1) = √1 =1

The range is:

0 < y < 1

4 0
3 years ago
How to find the average squared distance between the points of the unit disk and the point (1,1)
ddd [48]
The unit disk can be parameterized by the function

\mathbf p(r,\theta)=(r\cos\theta,r\sin\theta)

where 0\le r\le 1 and 0\le\theta\le2\pi. The squared distance between any point in this region (x,y)=(r\cos\theta,r\sin\theta) and the point (1, 1) is

(x-1)^2+(y-1)^2=(r\cos\theta-1)^2+(r\sin\theta-1)^2
=(r^2\cos^2\theta-2r\cos\theta+1)+(r^2\sin^2\theta-2r\sin\theta+1)
=r^2(\cos^2\theta+\sin^2\theta)-2r(\cos\theta-\sin\theta)+2
=r^2-2r(\cos\theta-\sin\theta)+2

The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.

\dfrac{\displaystyle\iint_{x^2+y^2
=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}
=\dfrac{\frac{5\pi}2}\pi=\dfrac52
5 0
3 years ago
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