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Vika [28.1K]
3 years ago
10

Solve the inequality. 4│x + 5│ - 2 ≤ 10

Mathematics
2 answers:
castortr0y [4]3 years ago
6 0

Step-by-step explanation: To solve this absolute value inequality,

our goal is to get the absolute value by itself on one side of the inequality.

So start by adding 2 to both sides and we have 4|x + 5| ≤ 12.

Now divide both sides by 3 and we have |x + 5| ≤ 3.

Now the the absolute value is isolated, we can split this up.

The first inequality will look exactly like the one

we have right now except for the absolute value.

For the second one, we flip the sign and change the 3 to a negative.

So we have x + 5 ≤ 3 or x + 5 ≥ -3.

Solving each inequality from here, we have x ≤ -2 or x ≥ -8.

kati45 [8]3 years ago
3 0

Answer:

4x+20 -2<10

4x+18+10

4x+28

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liq [111]

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Answer:

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Step-by-step explanation:

A plane is named either by its designator (Q) or by three non-collinear points in the plane. Here, the point I must be used, as the other designated points in the plane are all on the same line. The appropriate choice is ...

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4 0
2 years ago
(m? - 5 m +6 )<br>÷(m - 2)​
leonid [27]

Answer:

(m? - 5 m +6 )  ÷ (m - 2)​ = \frac{m}{-5m+6} / (m - 2)

Step-by-step explanation:

Divide the first expression by the second expression.

\frac{m}{-5m+6} / (m - 2)

I hope this helps.

4 0
3 years ago
Read 2 more answers
Use properties of operations to write an expression equivalent to the sum of the expressions 3(g + 5) and 2(3g - 6)
melisa1 [442]
Well,  you would istribute
using
a(b+c)=ab+ac
so

3(g+5)=3(g)+3(5)=3g+15
2(3g-6)=2(3g)+2(-6)=6g-12

now we gots
3g+15+6g-12
using commutative, we can arrange and group like terms because
a+b=b+a
so
3g+6g+15-12
add
9g+3
3 0
3 years ago
What is the quotient (2x^3 + 3x - 22) / (x-2)
Scilla [17]

Answer:

The quotient is 2x^2+4x+11

Step-by-step explanation:

6 0
3 years ago
find the slope of the curve y=x^2-2x-5 at the point P(2,5) by finding the limit of secant slopes through point P
Fynjy0 [20]

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, (t,y(t))=(t,t^2-2t-5), where t>2. The slope of the secant line through P and Q is given by the difference quotient,

\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t

where we are allowed to simplify because t\neq2.

Then the equation of the secant line is

y-(-5)=t(x-2)\implies y=t(x-2)-5

Taking the limit as t\to2, we have

\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

\dfrac{\mathrm dy}{\mathrm dx}=2x-2

and at x=2, we get a slope of 2(2)-2=2, as expected.

4 0
3 years ago
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