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Lena [83]
3 years ago
9

I need an equation that equals -5

Mathematics
2 answers:
RoseWind [281]3 years ago
8 0
-4 - 1 = -5 can be an equation that equals -5.
lyudmila [28]3 years ago
6 0

Answer:

-2+-3 is an equation that would fit

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arsen [322]
2.96 or 18.75 is the answer im leaning towards 2.96 though
5 0
3 years ago
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What are the end points of the triangle called? How are they labeled?
horrorfan [7]

For this case we have that by definition:

A triangle is defined by three lines that are called sides, or by three points called vertices.

We know that:

  1. <em>The vertices of a triangle are labeled with capital letters. </em>
  2. <em>The sides of a triangle are written with lowercase letters. </em>
  3. <em>The angles of a triangle are written similarly to the vertices. </em>

Answer:

The end points of the triangle are called vertices.

They are labeled in capital letters.

6 0
3 years ago
A positive integer is 5 more than 23 times another. Their product is 6732. Find the two integers.
inna [77]
Another way to write this is:
x = 23y + 5
xy = 6732

Now plug in the first equation into the second:
(23y + 5)y = 6732
23y^2 + 5y - 6732 = 0

Either use quadratic equation or factor:
(y - 17)(23y + 396) = 0
y = 17 or -396/23

You know you can automatically eliminate the second y because it's negative and you need the two integers to multiply to a positive number (6732).

Plug y = 17 back into either equation (second might be easier):
17x = 6732
x = 396

8 0
3 years ago
What is the key difference between the graph of a linear relationship and the graph of a nonlinear relationship?
kakasveta [241]
<span>The main key difference between the graph of a linear relationship and the graph of a nonlinear relationship are linear relationship is the relation between variables which creates a straight line when spotted on a cartesian plane and linear relations have constant slope always.The key difference between the graph of an exponential relationship and the graph of a quadratic relationship is exponential relation is a mathematical function of the following form: f ( x ) = a x. where x is a variable, and a is a constant called the base of the function but quadratic relationship of the graph is the the standardized form of a quadratic equation is ax^2 + bx + c = 0,.</span>
4 0
3 years ago
Read 2 more answers
What is the antiderivative of 3x/((x-1)^2)
Maslowich

Answer:

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Step-by-step explanation:

Given

\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx

\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx

=3\cdot \int \frac{x}{\left(x-1\right)^2}dx

\mathrm{Apply\:u-substitution:}\:u=x-1

=3\cdot \int \frac{u+1}{u^2}du

\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}

=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du

\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx

=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)

as

\int \frac{1}{u}du=\ln \left|u\right|     ∵ \mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)

\int \frac{1}{u^2}du=-\frac{1}{u}        ∵     \mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1

so

=3\left(\ln \left|u\right|-\frac{1}{u}\right)

\mathrm{Substitute\:back}\:u=x-1

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)

\mathrm{Add\:a\:constant\:to\:the\:solution}

=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

Therefore,

\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C

4 0
3 years ago
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