All categories

4 years ago

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

You might be interested in

Mr. Franklin asked each of his 30 students to select their favorite radio station.

Ostrovityanka [42]

Answer:

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:

idek

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.