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3 years ago

after 5 years of earning at an annual rate of 4 percent, an investment has earned 1200 in interest. determine the amount of the

initial investement

Mathematics

1 answer:

8_murik_8 [283]3 years ago

3 0

$\bf ~~~~~~ \textit{Simple Interest Earned}
\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\to &\$1200\\
P=\textit{original amount deposited}\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
t=years\to &5
\end{cases}
\\\\\\
1200=P(0.04)(5)\implies \cfrac{1200}{(0.04)(5)}=P\implies 6000=P$

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Saylor Henshaw earned $120 in simple interest in 9 months at an annual interest rate of 5%. How much money did he invest?

Ksju [112]

Annual interest is P*r^t, r is 0.05 here, t is 9/12=3/4=0.75
P*0.05^0.75=120
use your calculaor: P= 120÷ (0.05)^0.75=1135
so he invested about 1135 dollars

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3 years ago

Select all the expressions that are equivalent to (2)^n+³

eimsori [14]

Answer:

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

$a^{m}$ × $a^{m}$ = $a^{m+n}$
$(a^{m})^{n}$ = $a^{m*n}$

Now let us find the equivalent expressions of $(2)^{n+3}$

∵ 4 = 2 × 2

∴ 4 = $2^{2}$

∴ $(4)^{n+2}$ = $(2^{2})^{n+2}$

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴ $(4)^{n+2}$ = $(2)^{2n+4}$

∵ 4 = 2 × 2

∴ 4 = 2²

∴ $4(2)^{n+1}$ = 2² × $(2)^{n+1}$

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴ $4(2)^{n+1}$ = $(2)^{n+3}$

∵ 8 = 2 × 2 × 2

∴ 8 = 2³

∴ $8(2)^{n}$ = 2³ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴ $8(2)^{n}$ = $(2)^{n+3}$

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = $2^{4}$

∴ $16(2)^{n}$ = $2^{4}$ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴ $16(2)^{n}$ = $(2)^{n+4}$

$(2)^{2n+3}$ is in its simplest form

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

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3 years ago

PLEASE SOMEONE HELP ILL GIVE BRAINLIEST!

grin007 [14]

Answer:

Part A

The rotation of the point (x, y) 180° about the origin gives the point (-x, -y)

Therefore, we have the points, ΔLMN, L(1, 1), and M(2, 2), and N(3, 3)

The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)

Please find attached the graph of triangle ΔLMN and ΔL'M'N'

Part B

The lines drawn through L and L' and through M and M' are colinear

Part C

A line is defined by two points, such as L and M. Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.

The third point, N, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage

Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'

Step-by-step explanation: