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adoni [48]
4 years ago
9

What is 60 times 20 plz answer and make it quick!!

Mathematics
2 answers:
timurjin [86]4 years ago
6 0
1200 hope this helps u
Veronika [31]4 years ago
4 0
60 × 20= 1200

This time there are 2 methods:

1st method: Pretend that both zeros are gone and you are left with 6 × 2= 12
Then you could just add the 2 zeros that you pretended that were gone.

2nd method: If you want to show work:

 60
 x
 20
------
 000
+
120
--------
1200
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Mr. Franklin asked each of his 30 students to select their favorite radio station.
Ostrovityanka [42]

Answer:

14

Step-by-step explanation:

We are given that

Total number of students who select their favorite radio station=30

Number of student who select their favorite radio station K=\frac{1}{3}\times 30=10

Number of student who select their favorite radio station L=\frac{1}{5}\times 30=6

We have to find that number of student who select station M.

Remaining students=30-(10+6)=30-16=14

Number of students who select their favorite radio station M=14

Hence, the number of students who select their favorite radio station M=14

Answer: 14

6 0
3 years ago
Help me with question 4 and 5
lisov135 [29]

Answer:

2.6768·10^(-14)  (Answer C)

Step-by-step explanation:

Problem 4:  We ADD the powers of 10:  10^6·10^(-3) = 10^3.  Also, 3.6·1.7 = 6.12.  Therefore, Answer D is correct.

Problem 5:  Again we ADD (combine) the powers of 10:  10^(-10)·10^(-5) =

10^(-15).  4.78·5.6 = 26.768.  Our intermediary answer is thus:

26.768·10^(-15).  This must be rewritten in the form x.xxxx, so here we have:

2.6768·10^(-14)  (Answer C)

3 0
4 years ago
F(x)=3x+5, g(x)=6x^2 find (fg)(x)
alexgriva [62]

Answer:

18x^3+30x^2

Step-by-step explanation:

im smart.

6 0
3 years ago
PLS HELP ASAP IS THIS TRUE OR FALSE?<br><br> *image below*
Darya [45]

Answer:

true

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Please help with these partial fractions!!!
VARVARA [1.3K]

a. Factorize the denominator:

\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}

Then we're looking for a,b such that

\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}

\implies x+14=a(x+2)+b(x-4)

If x=4, then 18=6a\implies a=3; if x=-2, then 12=-6b\implies b=-2. So we have

\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}

as required.

b. Same setup as in (a):

\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}

We want to find a,b,c such that

\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant b' such that the numerator of the second term is b'x+b. We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining b'=0 anyway.

\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2

If x=0, then b=6; if x=-1, then c=-2. Expanding everything on the right then gives

-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6

which tells us a-2=-3 and a+6=5; in both cases, we get a=-1. Then

\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}

as required.

5 0
4 years ago
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