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2 years ago

How much interest is earned on $8,000 invested at an annual rate of 12% for 1 year?

Mathematics

1 answer:

Ghella [55]2 years ago

8 0

Answer:

96$

Step-by-step explanation:

P = $8,000

R = 12%

T = 1yr

I = PRT/100

I = (800×12×1)/100

I = 96$

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Can yall help me with the second part Pls

miskamm [114]

Answer:

Step-by-step explanation:

Total money raised = 658 + 95.60 = $ 753.60

donated to each charity = 753.60/12 = $62.80

7 0

3 years ago

A tipping platform is a ramp used to unload trucks. How high is the end of an 100 foot ramp when it is tipped by a 30 degree ang

Naily [24]

According to the information given, the ramp would be 50 feet high. (However, that seems really high. Did you type your problem correctly?)

If you draw a picture, you will see that you can set up a cosine ratio to solve this problem. With a 30 degree angle of depression, the angle we are going to use 60 degrees.

We can set up the following equation:

cos(60) = x / 100
0.5 = x / 100
50 = x

6 0

3 years ago

Will give brainliest if correct.

Nataly [62]

Answer:

The answer would be, 81cm.

Step-by-step explanation:

36÷4=9

9×9=81

8 0

3 years ago

Read 2 more answers

Joshua invested $750 in an account paying an interest rate of

Maslowich

Due to the difference in the interest rate and the quarterly compounding, Joshua will have $212.24 more than Josiah.
Step-by-step explanation:
Giving the following information:
Joshua:
Initial investment (PV)= $750
Interest rate (i)= 0.0341/4= 0.008525
Number of periods (n)= 18*4= 72 quarters
Josiah:
Initial investment (PV)= $750
Interest rate (i)= 0.0285
Number of periods (n)= 18 years
To calculate the future value of each one, we need to use the following formula:
FV= PV*(1 + i)^n
Joshua:
FV= 750*(1.008525^72)
FV= $1,381.98
Josiah:
FV= 750*(1.0285^18)
FV= $1,169.74
Due to the difference in the interest rate and the quarterly compounding, Joshua will have $212.24 more than Josiah.

5 0

2 years ago

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

2 years ago