All categories

3 years ago

Which compound inequality can be used to solve the inequality 13x+2 >7?

Mathematics

1 answer:

VashaNatasha [74]3 years ago

4 0

Answer:

(D) $3x+2 < -7$ or $3x+2 >7$

Step-by-step explanation:

Given the absolute inequality: $|3x+2| >7$

When solving absolute inequalities, if the problem has a greater than sign we set up an "OR" compound inequality that looks like this:

(Expression inside absolute value) < - (number on other side) OR
(Expression inside absolute value) > (number on other side)

Therefore, for the absolute inequality |3x+2| >7, we have:

$3x+2 < -7$ or $3x+2 >7$

The correct option is D.

You might be interested in

System of equations -8x + 4y = 24 -7x + 7y = 28 X= ___ Y=___

lilavasa [31]

Answer:

You would end up with X= -2, y= 2

And they would intersect at point (-2,2) on a graph :)

8 0

3 years ago

Read 2 more answers

Find the exact value of cos(arcsin(1/4)). For full credit, explain your reasoning.

Westkost [7]

First way
arcsin(1/4) means that 1/4 sin of the angle.

sin(α)=1/4
sin²α+cos²α=1
(1/4)²+cos²α=1
cos²α=1-1/16 =15/16
cosα=+/-(√15)/4

Second way

sin(α)=1/4 =opposite leg/hipotenuse
cos(α)=adjacent leg/hypothenuse
adjacent leg =√(hypotenuse²- Opposite²)=√(16-1)=√15
cosα=+/-√15/4

For one value of sinα, possible 2 values of cosα.

5 0

3 years ago

an amusement park charge 35.50 for admission.on saturday,6,789 people visited the park.about how much money did the park earn fr

sp2606 [1]

6,789
-
35.50
______
32.39

3 0

3 years ago

Read 2 more answers

If u=<2,2> then find u•u and determine whether it is a vector or a scalar

Kitty [74]

If ~v = hv1, v2, v3i and ~w = hw1, w2, w3i are vectors and c is a scalar, then (a) c~v = hcv1, cv2, cv3i (b) ~v + ~w = hv1 + w1, v2 + w2, v3 + w3i (c) ~v − ~w = hv1 − w1, v2 − w2, v3 − w3i.

3 0

3 years ago

Log(x-2)+log2=2logy log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

3 years ago