MATHEMATICS MIDDLE SCHOOL

What is the value of 2 in 5,014,245?I wrote 200 and got it wrong. The 2 is in the hundreds place

Answers

Answer 1
Answer: Your reasoning and your answer were both correct. Discuss with the teacher. ("Discuss". DON'T fight.)

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Can u help plz because i dont get it!!!!12 The price of a cake is £C.
The total cost of buying 20 cakes is £T.
Write a formula for T in terms of C.
thanks
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Answers

so the price is C
and 20 cakes is T

So 1 cake times 20=20 cakes
so 20C=T

How don I simplify 18/48

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You simplify it by dividing 18/48 by 2 and keep simplifying till you get your final results.

18/48 = 3/8 


You find out what number they both have in common. So, in this case

18/48

They are both divisible by 6 (6 is the greatest common factor between the two.)

Now ,divide 6 and 8 then 6 and 48 to get

3/8 

If a school bought 1,000 erasers as part of an order for supplies. The total cost of the erasers was $30. What was the cost for 1 eraser?

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The cost of 1000 erasers was 30$, then one eraser would be 30/1000=0.03$

-4.3g=25.8 what is the awnser

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-4.3g=25.8\ \ \ \ |divide\ both\ sides\ by\ (-4.3)\n\ng=(25.8)/(-4.3)\n\ng=-(258)/(43)\n\n\huge\boxed{g=-6}
\sf - 4.3g = 25.8 \n \n Divide \ both \ sides \ by \ - 4.3 \n \n (- 4.3g)/(- 4.3) = (25.8)/(-4.3) \n \n \huge = g = - 6

Solve.
8+h>2+3h
A) h>3
B) h>4
C) h<3
D) h<4

Answers

8 + h > 2 + 3h
h - 3h > 2 - 8 . . . . . . . . (collecting like terms)
-2h > -6
h < -6/-2 . . . . . . . . (> changes to < when you divide with negative number and vice-versa)
h < 3 (i.e. the third option)

Some help me with this please!!

Answers

First of all observe that

r^6s^3=(r^2s)^3 \implies \sqrt[3]{r^6s^3}=\sqrt[3]{(r^2s)^3}=r^2s

So, we have

\log(r^5s^6\sqrt[3]{r^6s^3})=\log(r^5s^6\cdot r^2s)

We can simplify the input as

\log(r^5s^6\cdot r^2s)=\log(r^7s^7)

The logarithm of a multiplication is the sum of the logarithms:

\log(r^7s^7)=\log(r^7)+\log(s^7)

Finally, invoke the logarithm property \log(a^b)=b\log(a)

to get the final answer

\log(r^7)+\log(s^7)=7\log(r)+7\log(s)
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