All categories

3 years ago

Find the total amount in the compound interest account.

Mathematics

1 answer:

ivanzaharov [21]3 years ago

5 0

11340

9% of 7000 is 630 then just multiply 630x18 which equals to 11340

You might be interested in

What is the measure, in degrees, of an angle that represents 50/350 of a circle?

JulsSmile [24]

Answer:

$\mathrm{A.\: 50^{\circ} }$

Step-by-step explanation:

Since there are 360 degrees in a circle, we can set up the following proportion:

$\frac{50}{360}=\frac{x}{360},\\360x=50\cdot 360,\\x=\boxed{50}$

8 0

3 years ago

Mr. Clark traveled 456.4 km in 14 days.

xeze [42]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

<h3><u>Distance travelled each day is</u> : </h3>

32.6 km

<h3><u>Explanation</u> : </h3>

Distance travelled in 14 days is : 456.4 km

And if he travels same Distance each day, so let the Distance travelled on each day be x

According to question ~

$\qquad \sf \dashrightarrow \: 14 \times x = 456.4$

$\qquad \sf \dashrightarrow \: x = 456.4 \div 14$

$\qquad \sf \dashrightarrow x = \: 32.6 \: \: km$

So, he travels 32.6 km per day

5 0

2 years ago

Will give brainliest to right answer Homies

BigorU [14]

Answer:

C :) hope it helps

Step-by-step explanation:

3 0

3 years ago

PLEASE HELP ME PLEAESESES

Bogdan [553]

Answer:

Make a coordinate plane and you have to solve around the orgin say 4,4 and 4,1

Step-by-step explanation:

8 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago