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3 years ago

14

Mitchell travels from the US to Canada, where he exchanges 150 US dollars for Canadian dollars. He then spends 20 Canadian dolla

rs, returns to the US, and exchanges the remaining money back to US dollars. How many US dollars does Mitchell have remaining?

2 answers:

Katarina [22]3 years ago

8 0

Answer:C 130.66

Step-bcc-step explanation:

lora16 [44]3 years ago

3 0

Their money rate is the same as ours so $150- 20= $ !30

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I am very confused, please help!

mojhsa [17]

Answer:

D

Step-by-step explanation:

Consider the equation:

$\dfrac{3}{4}(24-16x)=-\dfrac{2}{3}(18x-27)$

Use distributive property in both sides:

$\dfrac{3}{4}\cdot 24-\dfrac{3}{4}\cdot 16x=-\dfrac{2}{3}\cdot 18x+\dfrac{2}{3}\cdot 27\\ \\3\cdot 6-3\cdot 4x=-2\cdot 6x+2\cdot 9\\ \\18-12x=-12x+18$

Separate terms with x into the left part and terms without x into the right part:

$-12x+12x=18-18\\ \\0=0$

Since you get true equality 0 = 0, the equation is true for all real values of x. So, the equation has infinitely many solutions.

6 0

3 years ago

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

As

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

3 years ago

A short necklace has 32 gold beads and 8 black beads.

Katena32 [7]

I think there are 12
so there are 40 beads on the short one
and 60 on the long one so i multiplied 8*1.5 and got 12

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3 years ago

Simplify the expression:<br><br> –5(6 − 2y) =

Vadim26 [7]

-30+10y
hope this helps!

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3 years ago

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Recall that you can use the built-in functions rem(m,n) or mod(m,n)to find the remainder of the division m/n.

suter [353]

The answer is B hope this helped

5 0

3 years ago