a man invested a total of $3,000 in two investments. he made a profit of 3% on the first investment and 4% on the second investment. if his total profit was $107, what was the amount of each investment?

Answers

Answer 1

Answer: Let the amount that the man placed in the first investment be $x$ . Then, the amount that he placed in the second investment has to be $\$3000 - x$ . Using those as the investment amounts, the total profit is given by adding the two separate profits as shown:

$.03x + .04(\$3000 - x) = \$107$

We can now solve for x:

$.03x + .04(\$3000 - x) = \$107$
$.03x + (\$120 - .04x) = \$107$
$\$120 - .01x = \$107$
$.01x = \$13$
$x = \$1300$

Thus,

First investment: $x = \bf \$1300$
Second investment: $\$3000 - x = \$3000 - \$1300 = \bf \$1700$

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Which fractions are equal to
2
3
?

Answers

Answer:

the answer would be 4/6

Step-by-step explanation:

because u multipy 2 by 2 and 3 by 2, then get 4/6

Using the formula C = 2πr, find the circumference of a circle with diameter of 28 inches. Round your answer to the nearest inch. A. 28 in. B. 56 in. C. 44 in. D. 88 in.

Answers

I hope this helps you

175.92918845 is what i got so idk i am sorry

The yearbook club had a meeting. The meeting had 15 people, which is three-fifths of the club. How many people are in the club?

Answers

25.
For me, I just did 15 divided by 3, which equals to 5. So.. You do 5 x 5 to find out how many people there are in the club in total.

Final answer:

The question is about finding the total number of people in the yearbook club given that 15 people represent three-fifths of the total. By using a proportion, we find that the yearbook club has 25 people.

Explanation:

This is a proportional relationship problem in mathematics. You are given that 15 people represent three-fifths (or 3/5) of the total number of people in the yearbook club. To find the total number of people in the club, you set up the proportion: 15 is to 3 and X is to 5.

Then, cross multiply and solve the equation for X. 3*X = 15*5, therefore X = 75/3 = 25. So, there are 25 people in the yearbook club.

Learn more about Proportional Relationships here:

brainly.com/question/34138295

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Trig help (Problems already solved)?The calculation for the problems are already done but I have to list a reason or what is being done in each step. "Each = and newline made" means a place I have to write what is being done in the calculation.

1. (secx + sinx)cotx = cscx + cosx
=(secx + sinx)cotx = cscx + cosx
=(1 / sinx) + cosx
=cscx + cosx

2. cosx + tanx sinx = secx
=cosx + tanx sinx = cosx + (sinx / cosx)sinx
=cosx + (sin^2x / cosx) = (1 / cosx)(cos^2x + sin^2x)
=1 / cosx
=secx

3. cscx - cosx cotx = sinx
=cscx - cosx cotx = (1 / sinx) - cosx(cosx / sinx)
=(1 / sinx) - (cos^2x / sinx)
=(1 - cos^2x) / sinx
=sin^2x / sinx = sinx

4. (cosx / (1 + cosx)) + (cosx / (1 - cosx)) = 2cotx cscx
=(cosx / (1 + cosx)) + (cosx / (1 - cosx)) = ((cosx (1 - cosx) + cosx (1 + cosx))) / (1 + cosx)(1 - cosx)
=(cosx - cos^2x + cosx + cos^2x) / (1 - cos^2x)
=2cosx / sin^2x
=2(cosx / sinx)(1 / sinx) = 2cotx cscx

Thank you to whoever decides to help me with explaining what is happening on each line.

Answers

The first identity uses the definition of the reciprocal functions $\sec x,\csc x,\cot x$ and the distributive property of multiplication.

$(\sec x+\sin x)\cot x=\left(\frac1{\cos x}+\sin x\right)(\cos x)/(\sin x)$
$=(\cos x)/(\cos x\sin x)+(\cos x\sin x)/(\sin x)$
$=\frac1{\sin x}+\cos x$
$=\csc x+\cos x$

The second uses the definition of $\tan x$ and the distributive property. Then a factor of $\frac1{\cos x}$ is pulled out, which allows you to use the identity $\sin^2x+\cos^2x=1$ .

$\cos x+\tan x\sin x=\cos x+(\sin x)/(\cos x)\sin x$
$=\cos x+(\sin^2x)/(\cos x)$
$=(\cos^2x)/(\cos x)+(\sin^2x)/(\cos x)$
$=\frac1{\cos x}\left(\cos^2x+\sin^2x\right)$
$=\frac1{\cos x}*1$
$=\frac1{\cos x}$
$=1$

The third uses the same ideas as the second: rewrite the reciprocal functions, then invoke the Pythagorean identity $\sin^2x+\cos^2x=1$ , which is equivalent to $\sin^2x=1-\cos^2x$ .

$\csc x-\cos x\cot x=\frac1{\sin x}-\cos x(\cos x)/(\sin x)$
$=\frac1{\sin x}-(\cos^2x)/(\sin x)$
$=\frac1{\sin x}\left(1-\cos^2x\right)$
$=\frac1{\sin x}\sin^2x$
$=(\sin^2x)/(\sin x)$
$=\sin x$

In the last one, you combine the fractions by enforcing common denominators. This lets you add the numerators together, and the denominator can be simplified. Once you do that, you rewrite the factors of cos and sin in the numerator and denominator to make up the cot and csc functions, and you're done.

$(\cos x)/(1+\cos x)+(\cos x)/(1-\cos x)=(\cos x(1-\cos x))/((1+\cos x)(1-\cos x))+(\cos x(1+\cos x))/((1-\cos x)(1+\cos x))$
$=(\cos x(1-\cos x)+\cos x(1+\cos x))/((1-\cos x)(1+\cos x))$
$=(\cos x(1-\cos x+1+\cos x))/(1-\cos^2x)$
$=(2\cos x)/(\sin^2x)$
$=2(\cos x)/(\sin x)\frac1{\sin x}$
$=2\cot x\csc x$

Triangle PQR ~ triangle PꞌꞌQꞌꞌRꞌꞌ.Which two transformations were performed on triangle PQR to produce triangle PꞌꞌQꞌꞌRꞌꞌ?

Choose exactly two answers that are correct.

A.Triangle PQR was reflected across the y-axis to produce triangle PꞌQꞌRꞌ.

B.Triangle PQR was reflected across the x-axis to produce triangle PꞌQꞌRꞌ.

C.Triangle PꞌQꞌRꞌ was dilated by a scale factor of to produce triangle PꞌꞌQꞌꞌRꞌꞌ.

D.Triangle PꞌQꞌRꞌ was dilated by a scale factor of 2 to produce triangle PꞌꞌQꞌꞌRꞌꞌ.

Answers

The correct answer for the question that is being presented above is this one: D.Triangle PꞌQꞌRꞌ was dilated by a scale factor of 2 to produce triangle PꞌꞌQꞌꞌRꞌꞌ; C.Triangle PꞌQꞌRꞌ was dilated by a scale factor of to produce triangle PꞌꞌQꞌꞌRꞌꞌ.

A furniture maker uses the specification 19.88 ≤ w ≤ 20.12 for the width w in inches of a desk drawer. Write the specification as an absolute value inequality.

Answers

A furniture maker uses the specification 19.88 ≤ w ≤ 20.12

The absolute value inequality is

$|x-20|\leq 0.12$

Given :

A furniture maker uses the specification 19.88 ≤ w ≤ 20.12 for the width w in inches of a desk drawer

We need to write the given inequality in absolute value inequality

if $a-b<x<a+b$ then absolute value inequality is

$|x-a|< b$

To find out value of 'a' and 'b' we need to use the given inequality

compare a-b<x<a+b with given inequality

$a-b=19.88\na+b=20.12$

Solve for 'a' and 'b'

Add both equations

$2a=40\na=20$

Now find out b

$a+b=20.12\n20+b=20.12\nb=20.12-20\nb=0.12$

The required absolute value inequality is

$|x-a|\leq b\n|x-20|\leq 0.12$

Learn more : brainly.com/question/1770168

The correct answer is:

|w-20| ≤ 0.12.

Explanation:

We first find the average of the two ends of the inequality:

(19.88+20.12)/2 = 40/2 = 20

This will be the number subtracted from w in the inequality.

Now we find the difference between this value and the ends:

20-19.88 = 0.12

20.12 - 20 = 0.12

This will be what our absolute value inequality ends with; the "answer" part, so to speak.

Since this inequality is written in compact form, it must be an "and" inequality; this means the absolute value inequality must be a "less than or equal to."

This gives us

|w-20| ≤ 0.12

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