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Norma-Jean [14]
3 years ago
6

Find last year’s salary if, after a 3% pay raise, this year’s salary is $32,960.

Mathematics
1 answer:
belka [17]3 years ago
3 0

Answer:

$32000

Step-by-step explanation:

Let last years salary was $x

Therefore,

x + 3% of x = $32960

x + 0.03x = $32960

1.03x=$32960

x = 32960/1.03

x = $32,000

So, last year's salary was $32000.

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Answer the precalculus question from khan academy pls! I will give brainliest and 5 stars. 30 points.
rosijanka [135]

The value of f(h(2)) =2 and h(f(16))= 14

<h3>What is function?</h3>

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input . Mapping or transformation is used to denote a function in math. These functions are usually denoted by letters . The domain is defined as the set of all the values that the function can input while it can be defined. The range is all the values that come out as the output of the function involved.

given:

f(x)= √x-1 , h(x)= x² + 5

Now,

f(h(2))= f( (2)² +5 )

=f(4+5)

=f(9)

=√9-1

= 3-1

=2

h(f(16)) = h( √16-1)

=h( 4-1)

=h(3)

=3² + 5

=9+5

=14

Learn more about function here:

brainly.com/question/12431044

#SPJ1

7 0
2 years ago
Let q(x)=8x-7 what is q(4) a. 25 b. 36 c. 39 d. 32
zhuklara [117]
I hope this helps you



x=4 q (4)=8.4-7


q (4)=32-7


q (4)=25
4 0
3 years ago
Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?
Yanka [14]

Answer:

Step-by-step explanation:

\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.

\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\

(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+  \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.

8 0
1 year ago
Timothy has $1111 and Mary has $777. How much money does Timothy need to give Mary, so that they would have the same amount of m
ZanzabumX [31]

Answer:

167

We do 1111+777 for the total and that is 1888

1888/2=944

then 1111-944=167; then they will have equal amounts!!

LUV YA!!

3 0
3 years ago
Can someone pls answer this?❤️
nata0808 [166]

Answer:

i want to say the answer is A

Step-by-step explanation:

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