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Anastaziya [24]

3 years ago

9

The product of the slopes of perpendicular lines is always

1 answer:

erastovalidia [21]3 years ago

8 0

The product of the slopes is -1.

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Can someone plz explain how to do this?

andre [41]

Subtract y
3x - y + 5 = 0

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3 years ago

PLEASE HELP I GIVE THANKS

GenaCL600 [577]

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4 years ago

An art teacher has 70 1/5 pounds of clay and 20 students. Each student will get an equal amount of clay. How much clay does each

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2 years ago

Suppose you deposit $2500 in a savings account that pays you 5% interest per year. (Calculator)

BARSIC [14]

Answer:

14.20669 years

Roughly 14 years and 2.5 months.

Step-by-step explanation:

Assuming this is compound interest.

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

$A=$ Final Amount

$P=$ Principal Amount

$r=$ Interest Rate

$n=$ # of times interest is compounded per year

$t=$ Time in years

We are looking for the times in years to double the money so

$2500*2=5000$

$A=5000$

$P=2500$

$r=0.05$

$n=1$

$t=?$

Lets solve for $t$ .

Step 1.

Plug in our numbers into the compound interest formula.

$5000=2500(1+\frac{0.05}{1}) ^{1*t}$

Step 2.

Simplify the equation.

Evaluate $1+\frac{0.05}{1}=1.05$

Evaluate $1*t=t$

$5000=2500(1.05) ^{t}$

Step 3.

Divide both sides of the equation by $2500$

$\frac{5000}{2500}=1.05 ^{t}$

Evaluate $\frac{5000}{2500}=2$

$2=1.05 ^{t}$

Step 4.

Take the natural log of both sides of the equation and rewrite the right side of the eqaution using properties of exponents/logarithms.

$ln(2)=t*ln(1.05)$

Step 5.

Divide both sides of the equation by $ln(1.05)$

$\frac{ln(2)}{ln(1.05)}=t$

Step 6.

Evaluate

$t=14.20669$

Roughly 14 years and 2.5 months.

4 0

2 years ago

Karleigh walks 5/8 mile to school every day.How far does she walk to school in 5 days

liberstina [14]

Answer:

$3\frac{1}{8}=3.125$ miles

Step-by-step explanation:

We have been given that Karleigh walks 5/8 mile to school every day.

To find the distance walked by Karleigh in 5 days we will multiply distance traveled in one day by total number of days.

$\text{Distance walked by Karleigh in 5 days}=\frac{5}{8}\times 5$

$\text{Distance walked by Karleigh in 5 days}=\frac{25}{8}$

$\text{Distance walked by Karleigh in 5 days}=3\frac{1}{8}=3.125$

Therefore, Karleigh walked $3\frac{1}{8}=3.125$ miles to school in 5 days.

4 0

3 years ago