All categories

2 years ago

Which is the product of 7/9 and 6? A. 1 4/9 B. 3 7/9 C. 4 2/3 D. 7 5/7

Mathematics

2 answers:

Vladimir79 [104]2 years ago

8 0

Answer:

C. 4 2/3

Step-by-step explanation:

1) Use this rule: a/b * c = ac/b.

7 * 6/9

2) Simplify 7 * 6 to 42.

42/9

3) Simplify.

14/3

4) Convert to mixed fractin form.

4 2/3

lora16 [44]2 years ago

4 0

Answer:

The product of 7/9 and 6:

$\\7/9 x 6 \\\\7x6=42\\\\42/9\\4 6/9\\\\4 is the whole number\\6/9 is the fraction that if we remove the 3 from both of us, it gives us the result 2/3\\ The answer is the B \\$

Step-by-step explanation:

the answer is the B

Because You only had to multiply 6 by 7 and the result, which would be 42 over 9, should be converted to a mixed number, which would be dividing 42 by 9, which is 4, and us over 6, which is four, which is the whole number, and 6 is the numerator and you don't change the nine since it is still the denominator.

Because the 4 is the whole number

And 6/9 is the fraction that if were move the 3 from both of us , it gives us the result is 2/3

So we have 4 2/3

the answer is the B

You might be interested in

A piece of equipment is purchased for $18,000 & has a salvage value of $2,000. The estimated life is 10 years & the meth

Travka [436]

Answer:

The total value would be 17,210.

8 0

3 years ago

Find the measure of the angle

Wewaii [24]

Answer:

Step-by-step explanation:

4 0

3 years ago

Radioactive Waste The rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt,

Dmitry_Shevchenko [17]

Answer:

$\displaystyle{\int^{\infty}_0 {50e^{-0.04t}}\, dt}=1250$

Step-by-step explanation:

Given:

$T =\displaystyle{\int^{\infty}_0 {Pe^{-kt}} \, dt}$

where,

T = total amount of waste

P = 50, the initial rate

k = 0.04

t = time

$T =\displaystyle{\int^{\infty}_0 {50e^{-0.04t}} \, dt}$

now we need to solve this integral!

$T =\displaystyle{50\int^{\infty}_0 {e^{-0.04t}} \, dt}$

$T = \left|50\left(\dfrac{e^{-0.04t}}{-0.04}\right)\right|^{\infty}_0$

$T = \left|-1250e^{-0.04t}\right|^{\infty}_0$

$T = (-1250e^{-0.04(\infty)})-(-1250e^{-0.04(0)})$

when any number has a power of negative infinity it is 0. because: $a^-{\infty} = \dfrac{1}{a^{\infty}} = \dfrac{1}{\infty} = 0$ , like something being divided by a very large number!