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2 years ago

Please help me solve 8n+2-6n-13=9+2n-13

Mathematics

2 answers:

qwelly [4]2 years ago

6 0

No solution since the 2n on both sides cross out

hoa [83]2 years ago

4 0

Answer:

It has no solution, unless there's a specific way it needs to be solved

Step-by-step explanation:

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Triangle MNP is dilated according to the rule DO,1.5 (x,y)Right arrow(1.5x, 1.5y) to create the image triangle M'N'P, which is n

Vikki [24]

Given:

The vertex of a triangle MNP are M(-4, 6), N(2, 6), and P(-1, 1).

The rule of dilation is:

$(x,y)\to (1.5x,1.5y)$

The image of triangle MNP after dilation is M'N'P'.

To find:

The coordinates of the endpoints of segment M'N'.

Solution:

The end points of MN are M(-4, 6) and N(2, 6).

The rule of dilation is:

$(x,y)\to (1.5x,1.5y)$

Using this rule, we get

$M(-4,6)\to M'(1.5(-4),1.5(6))$

$M(-4,6)\to M'(-6,9)$

And,

$N(2,6)\to N'(1.5(2),1.5(6))$

$N(2,6)\to N'(3,9)$

The endpoints of M'N' are M'(-6, 9) and N'(3, 9).

Therefore, the correct option is B.

8 0

3 years ago

(3.8 x 10 ^ -6) x (2.37 x 10 ^ -3)<br><br> Enter the answer in science notification

Black_prince [1.1K]

Multiply the coefficients and the powers of 10 with each other:

$(3.8 \times 10^{-6}) \times (2.37 \times 10^{-3}) = (3.8\times2.37) \times (10^{-6} \times10^{-3})$

The numeric part simply yields

$3.8\times2.37 = 9.006$

As for the powers of 10, you have to add the exponents, using the rule

$a^b \times a^c = a^{b+c}$

So, we have

$10^{-6} \times10^{-3} = 10^{-6-3} = 10^{-9}$

So, the final answer is

$9.006\times 10^{-9}$

7 0

2 years ago

Carlos Martin received a statement from his bank showing a balance of $56.75 as of March 15. His checkbook shows a balance of $8

kakasveta [241]

Checks not cancelled;
5.00+13.25
18.25
Total

7 0

3 years ago

Read 2 more answers

Simplify the following expression:

Bas_tet [7]

Answer:

3x-8y+40z

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Can you please help me out

Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

1 year ago