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3 years ago

Is is right? If not, please correct me (:

Mathematics

1 answer:

Illusion [34]3 years ago

6 0

Answer:

undefined

Step-by-step explanation:

The formula for slope is

m = (y2-y1)/ (x2-x1)

We have points

(-7,2) (-7,-7)

m = (-7-2) /(-7--7)

= -9/(-7+7)

= -9/0

This is undefined

We have points

(-8,-3) (6,-3)

m = (-3--3) /(6--8)

= (-3+3/(6+8)

= 0/14

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Find each percent change 40 to 2

Nostrana [21]

The question we need to ask is what percent of 2 is 40?

40 is 20 times 2 so 20 would be 2,000%

We need to subract 100% from any percent change:

2,000% - 100% = 1,900%

ANSWER: 1,900%

3 0

2 years ago

Helllp im doing my exam !

Tcecarenko [31]

log x ( 5 × 12 ) = 7

x^7 = 60

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3 years ago

When 2 fair dice are rolled there are 36 possible outcomes. How many possible outcomes would there be if three fair dice were ro

inysia [295]

216 as when 2 there is 6x6 but when there’s 3 it’s 6x6x6 which is 216

7 0

2 years ago

Read 2 more answers

Lou has an account with $10,000 which pays 6% interest compounded annually. If to that account, Lou deposits $5,000 at the begin

katovenus [111]

Answer:

Option d. $22154 is the right answer.

Step-by-step explanation:

To solve this question we will use the formula $A=P(1+\frac{r}{n})^{nt}$

In this formula A = amount after time t

P = principal amount

r = rate of interest

n = number of times interest gets compounded in a year

t = time

Now Lou has principal amount on the starting of first year = 10000+5000 = $15000

So for one year $A=15000(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 15000(1+.06)^{1}$

$= 15000(1.06)$ = $15900

After one year Lou added $5000 in this amount and we have to calculate the final amount he got

Now principal amount becomes $15900 + $ 5000 = $20900

Then putting the values again in the formula

$A=20900(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 20900(1+.06)^{1}$

$= 20900(1.06)=22154$

So the final amount will be $22154.

3 0

2 years ago

Rafael must choose a number between 67 and 113 that is a multiple of, 2,4, and 5 . Write all the numbers that he could choose.

skelet666 [1.2K]

First, find out what the greatest common factor is between 2, 4, and 5.

This is 20. So any number divisible by 20 is also divisible by 2, 4, and 5. 

Now we find the list of numbers that are multiples of 20 between 67 and 113. 

The solutions are: 

80 and 100.

5 0

3 years ago