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2 years ago

Light travels 1.18 x 10^10in. In 1 second. How far will light travel in 1 nanosecond or 1 x 10^-9s

Mathematics

1 answer:

alisha [4.7K]2 years ago

4 0

Make a formula then just use photomath lol.

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Can someone help me????

SpyIntel [72]

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5 0

2 years ago

I need help please

Levart [38]

The correct answer is true:)

7 0

2 years ago

Read 2 more answers

Suppose you'd like to save enough money to pay cash for your next car. The goal is to save an extra $28,000 over the next 6 year

Anna35 [415]

Given:

• Amount to save, A = $28,000

• Time, t = 6 years

• Interest rate, r = 5.3% ==> 0.053

• Number of times compounded = quarterly = 4 times

Let's find the amount that must be deposited into the account quarterly.

Apply the formula:

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

Where:

FV is the future value = $28,000

r = 0.053

n = 4

t = 6 years

Thus, we have:

$28000=P(\frac{(1+\frac{0.053}{4})^{4\times6}-1)}{\frac{0.053}{4}}$

Let's solve for P.

We have:

$\begin{gathered} 28000=P(\frac{(1+0.01325)^{24}-1}{0.01325}) \\ \\ 28000=P(\frac{(1.01325)^{24}-1)^{}}{0.01325}) \\ \\ 28000=P(\frac{1.371509114-1}{0.01325}) \\ \\ 28000=P(\frac{0.371509114}{0.01325}) \end{gathered}$

Solving further:

$28000=P(28.0384237)$

Divide both sides by 28.0384237:

$\begin{gathered} \frac{28000}{28.0384237}=\frac{P(28.0384237)}{28.0384237} \\ \\ 998.6=P \\ \\ P=998.6 \end{gathered}$

Therefore, the amount that must be deposited quarterly into the account is $998.60

ANSWER:

$998.60

8 0

5 months ago

OB is the bisector of AD and EC. . EO = 2x – 20 and OC = 70 – x. Solve for x.

mixas84 [53]

We know that OB bisects EC where they intersect would be at point O,
so the two lengths EO and OC must be equal.

2x - 20 = 70 - x

[adding x + 20 to each side]

2x - 20 + x + 20 = 70 - x + x + 20

[simplifying]

3x = 90

[dividing each side by 3]

x = 30 would be the answer

4 0

2 years ago

Find the coordinates of the midpoint of a segment with the endpoints (16,5) & (28,-13)

frutty [35]

$\text{The formula of the midpoint of a segment with the end points:}\\\\A(x_A;\ y_A)\ \text{and}\ B(x_B;\ y_B)\\\\M_{AB}=\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)$

$\text{We have:}\\A(16;\ 5)\to x_A=16;\ y_A=5\\B(28;\ -13)\to x_B=28;\ y_B=-13\\\\\text{substitute:}\\\\M_{AB}=\left(\dfrac{16+28}{2};\ \dfrac{5+(-13)}{2}\right)=\left(\dfrac{44}{2};\ \dfrac{-8}{2}\right)=(22;\ -4)$

7 0

2 years ago