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2 years ago

15

-2y-12 How do I show my work?????

1 answer:

grin007 [14]2 years ago

4 0

Factor - 2y - 12

- 2y - 12

= - 2( y + 6)

answer is - 2( y + 6)

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For which of the following inequalities is x≥4 a solution.<br> x+3≥1<br> 3x≥1<br> x-3≥1

pychu [463]

Answer:

x - 3 ≥ 1

Step-by-step explanation:

Solving each of the inequalities

x + 3 ≥ 1 ( subtract 3 from both sides )

x ≥ - 2

3x ≥ 1 ( divide each inequality by 3 )

x ≥ $\frac{1}{3}$

x - 3 ≥ 1 ( add 3 to both sides )

x ≥ 4 ← required solution

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2 years ago

Read 2 more answers

Pls help need ASAP Use the number line given to choose the best answer for the question.

DIA [1.3K]

Answer:

-3

Step-by-step explanation:

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3 years ago

Could someone answer and explain these please? Thank you!

Oksana_A [137]

Answer 1:

It is given that the positive 2 digit number is 'x' with tens digit 't' and units digit 'u'.

So the two digit number x is expressed as,

$x=(10 \times t)+(1 \times u)$

$x=10t+u$

The two digit number 'y' is obtained by reversing the digits of x.

So, $y=(10 \times u)+(1 \times t)$

$y=10u+t$

Now, the value of x-y is expressed as:

$x-y=(10t+u)-(10u+t)$

$x-y=10t+u-10u-t$

$x-y=9t-9u$

$x-y=9(t-u)$

So, $9(t-u)$ is equivalent to (x-y).

Answer 2:

It is given that the sum of infinite geometric series with first term 'a' and common ratio r<1 = $\frac{a}{1-r}$

Since, the sum of the given infinite geometric series = 200

Therefore, $\frac{a}{1-r}=200$

Since, r=0.15 (given)

$\frac{a}{1-0.15}=200$

$\frac{a}{0.85}=200$

$a=0.85 \times 200$

a=170

The nth term of geometric series is given by $ar^{n-1}$ .

So, second term of the series = $ar^{2-1}$ = ar

Second term = $170 \times 0.15$

= 25.5

So, the second term of the geometric series is 25.5

Step-by-step explanation:

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2 years ago

Which of the following is not a way to represent the solution of the inequality 8x – (5x + 4) 2-31?

Butoxors [25]

Answer:

caca sh

Step-by-step explanation:

shart

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2 years ago

Can someone pls pls PLS, help me out? ASAP!! (Geometry)<br> “Complete the proof”

Maru [420]

1) $\overline{AB} \perp \overline{BD}$ , $\overline{AB} \cong \overline{DB}$ , $\overline{AC} \cong \overline{DE}$ (given)

2) $\angle ABD$ is a right angle (perpendicular lines form right angles)

3) $\triangle ABD$ and $\triangle DEB$ are right triangles (a triangle with a right angle is a right angle)

4) $\triangle ABD \cong \triangle DEB$ (HL)

5) $\angle ACB \cong \angle DEB$ (CPCTC)

8 0

1 year ago