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3 years ago

Solve the compound inequality 7x > –35 and 3x ≤ 30.

Mathematics

2 answers:

dusya [7]3 years ago

8 0

Answer:-5 < x ≤ 10

Step-by-step explanation:

Ainat [17]3 years ago

7 0

<span>7x > -35
x > -35/7
x > -5

and 3x ≤ 30
x ≤ 30/3
x ≤ 10

put them together ..

-5 < x ≤ 10
</span>

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Graph the function <br> y=4(2^x+3)-1<br> and <br> y=-2(3^x-4)+1

kondaur [170]

The graphs of the functions y = 4(2ˣ + 3) - 1 and y = - 2(3ˣ - 4) + 1 are plotted on the graph.

<h3>What is function?</h3>

A function is a relation between a dependent and a independent variable, such that the dependent variable depends upon the independent one for its existence.

Given are the following functions -

y = 4(2ˣ + 3) - 1

y = - 2(3ˣ - 4) + 1

We have the following functions -

y = 4(2ˣ + 3) - 1

y = - 2(3ˣ - 4) + 1

Refer to the graph attached. The graph of red color represents the function y = 4(2ˣ + 3) - 1 and the graph of blue color represents the function y = - 2(3ˣ - 4) + 1. It can be seen that the graphs are symmetrical about the line y = 10.

Therefore, the graphs of the functions y = 4(2ˣ + 3) - 1 and y = - 2(3ˣ - 4) + 1 are plotted on the graph.

To solve more questions on plotting graphs, visit the link below-

brainly.com/question/14299685

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1 year ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$