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2 years ago

Solve this linear equation using substitution. (Please show work!)

Mathematics

2 answers:

julsineya [31]2 years ago

7 0

The answer is 7=1-7x

Genrish500 [490]2 years ago

5 0

Answer:

y=1-7x

Step-by-step explanation:

7x+y=1

The first thing to do is get the y by its self. So u cancel the 7x out by-7x and bring it to the other side.

7x-7x = 0

Now all we have left is: y=1-7x

Hope it helps and is right.

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144^2 - 121^2=

Mademuasel [1]

Answer:

Step-by-step explanation:

Use a² - b² = (a + b)(a - b)

144² - 121² = (144 + 121)(144 - 121)

7 0

3 years ago

A pair of sneakers are on sale for 20% off the original price. Tom purchased a pair of sneakers at this discounted price of $80.

LUCKY_DIMON [66]

The answer is 100.
You start by making a formula:
.80(x) = 80,
As we know the discount is 20% off, which makes the price actually 80% of what it originally was. Convert percent to a decimal gets you that .80
Reverse the process — 80/0.80 = 100

5 0

3 years ago

Read 2 more answers

Find the square root of 110 show your work

laila [671]

10.4880885
X 10.4880885 =110

4 0

3 years ago

a. Suppose we had $15,192 cash and invested it in the bank at 16 percent interest, how much would you have at the end of 1, 2, 3

Goryan [66]

Answer:

Part a) $\$17,622.72$

Part b) $\$20,442.36$

Part c) $\$23,713.13$

Part d) $\$27,507.23$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part a) How much would you have at the end of 1 year?

in this problem we have

$t=1\ years\\ P=\$15,192\\ r=0.16\\n=1$

substitute in the formula above

$A=15,192(1+\frac{0.16}{1})^{1*1}=\$17,622.72$

Part b) How much would you have at the end of 2 year?

in this problem we have

$t=2\ years\\ P=\$15,192\\ r=0.16\\n=1$

substitute in the formula above

$A=15,192(1+\frac{0.16}{1})^{1*2}=\$20,442.36$

Part c) How much would you have at the end of 3 year?

in this problem we have

$t=3\ years\\ P=\$15,192\\ r=0.16\\n=1$

substitute in the formula above

$A=15,192(1+\frac{0.16}{1})^{1*3}=\$23,713.13$

Part d) How much would you have at the end of 4 year?

in this problem we have

$t=4\ years\\ P=\$15,192\\ r=0.16\\n=1$

substitute in the formula above

$A=15,192(1+\frac{0.16}{1})^{1*4}=\$27,507.23$

5 0

3 years ago

Select all the sets of numbers that are possible values for x in the inequality, x<-3.

Solnce55 [7]

Answer: 2 and 3

Step-by-step explanation:

all (-9,-7,-5) and (-33,-22,-11) are negative numbers less than -3. So those answers are possible answers. ps i had this question on my test too.

6 0

3 years ago