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3 years ago

First question!!! Very simple but I don’t get it. WORTH 25 PTS

Mathematics

2 answers:

nexus9112 [7]3 years ago

6 0

Answer:

$\boxed{\sf m\angle RSV = {42}^{ \circ}}$

Given:

$\sf m\angle RST = 114^{\circ} \\ \\ \sf m\angle VST = 72^{\circ}$

To Find:

$\sf m\angle RSV$

Step-by-step explanation:

$\sf \implies m\angle RST = m\angle RSV + m\angle VST \\ \\ \sf \implies m\angle RSV = m\angle RST - m\angle VST \\ \\ \sf \implies m\angle RSV = {114}^{ \circ} - {72}^{ \circ} \\ \\ \sf \implies m\angle RSV = {42}^{ \circ}$

erica [24]3 years ago

4 0

Answer:

if the bottom the RST equals 114 you judt need to subtract 114 from 72 and you get 42 :) sorry if it wrong

Step-by-step explanation:

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Solve the equation-2(n+3)=13

bonufazy [111]

Answer:

im thinking n is 6.5 not sure wait not 6.5 its 3.5

7 0

2 years ago

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Robert is 10 years older than her brother. In 2 years, Robert will be three times as old as his brother. How old are they now?

valentina_108 [34]

Robert is 13 and his brother is 3 y.o.
X... brother’s age
X+10... Robert’s age
X+10+2... Robert’s age in 2 years
X+2...Brother’s age in 2 years
X+10+2/x+2=3
X=3
X+10=13

6 0

3 years ago

How much money will you have if you started with $1250 and put it in an account that earned 6.7% every year for 14 years?

KatRina [158]

Answer:

$3098.93

Step-by-step explanation:

We can use the formula for compound growth to solve this. The formula is:

$F=P(1+r)^t$

Where

F is the future value (the value at end of 14 years, our answer)

P is the initial amount invested ($1250)

r is the interest rate, in decimal (6.7% is 0.067)

t is the time in years (14, in our case)

<em>Plugging in all the information</em> we have:

$F=P(1+r)^t\\F=1250(1+0.067)^14\\F=1250(1.067)^14\\F=3098.93$

The account will accrue $3098.93 after 14 years.

3 0

3 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

notice how all the constants were combined into one big constant C.

7 0

3 years ago

(y - 2)2 = y2 – 6y + 4<br> Is this statement true or false?

trasher [3.6K]

<h3>Answer: False</h3>

==============================================

Explanation:

I'm assuming you meant to type out

(y-2)^2 = y^2-6y+4

This equation is not true for all real numbers because the left hand side expands out like so

(y-2)^2

(y-2)(y-2)

x(y-2) .... let x = y-2

xy-2x

y(x)-2(x)

y(y-2)-2(y-2) ... replace x with y-2

y^2-2y-2y+4

y^2-4y+4

So if the claim was (y-2)^2 = y^2-4y+4, then the claim would be true. However, the right hand side we're given doesn't match up with y^2-4y+4

--------------------------

Another approach is to pick some y value such as y = 2 to find that

(y-2)^2 = y^2-6y+4

(2-2)^2 = 2^2 - 6(2) + 4 .... plug in y = 2

0^2 = 2^2 - 6(2) + 4

0 = 4 - 6(2) + 4

0 = 4 - 12 + 4

0 = -4

We get a false statement. This is one counterexample showing the given equation is not true for all values of y.

6 0

2 years ago

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