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nevsk [136]
3 years ago
9

Do you have textbooks

Mathematics
1 answer:
Fittoniya [83]3 years ago
7 0

There are holt textbooks online if you have an Eclass course page.

You might be interested in
Which system of equations is incondsistent? a. 2x+8y=6 5x+20y=2
galina1969 [7]

Answer: Yes

Step-by-step explanation:

2x+8y=6

-8y -8y

2x = -2y

Divide by two so x will be by itself, so the answer will be -1y.

5x+20y=2

-20y -20y

5x = -18y

Divide by 5 so x will be by itself, so the answer will be -3y.

7 0
3 years ago
I need help please hurry.
Anarel [89]

Answer:the answer is the first

Step-by-step explanation:

Because the expression can be written as -10r2/2r2+4+5r/2r2+4+3/2r2+4

8 0
3 years ago
A zookeeper predicted that the weight of a newborn lion would be 2.8 pounds. When the zoo’s lion gave birth, the newborn weighed
Nonamiya [84]

3.5 - 2.8 is 0.7 so the zookeerper was 0.7 pounds off

5 0
4 years ago
Read 2 more answers
What is the value of this expression?
Mila [183]

Answer:

1/4

Step-by-step explanation:

1)2²= 4

2)4²= 16

3)3x2/3=2

4)6+2-4/8=1/4

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

6 0
3 years ago
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