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2 years ago

Maths 1. 24÷ (-6) 2. -15 ÷ 0.3 3. -4 ÷ (-20)

Mathematics

2 answers:

Lesechka [4]2 years ago

8 0

Answer:

hope it helps

Step-by-step explanation:

1) -72

2) -50

3) 0.2

Bas_tet [7]2 years ago

4 0

Answer:

1. 24÷ (-6)=− 4

Factor the numerator and denominator and cancel the common factors.

2. -15 ÷ 0.3=− 50

3. -4 ÷ (-20)=

Reduce the expression, if possible, by cancelling the common factors.

Exact Form:

1/5

Decimal Form:

0.2

here you go! :) hope I am right

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Answer:

#3: 84 books

#4: 90 pounds

Step-by-step explanation:

#3: 14*6= 84

#4: 30*3= 90

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2 years ago

Read 2 more answers

Given that f(x)= x2+3 and g(x)= x+4/3 solve for (f(g(x)) when x =2

Savatey [412]

$\bf \begin{cases}
f(x)=x^2+3\\\\
g(x)=x+\cfrac{4}{3}
\end{cases}
\\\\\\
\boxed{x=2}\qquad g(2)=(2)+\cfrac{4}{3}\implies \boxed{g(2)=\cfrac{10}{3}}
\\\\\\
f(~~g(x)~~)=[g(x)]^2+3\implies f(~~g(2)~~)=[g(2)]^2+3
\\\\\\
f(~~g(2)~~)=\left[ \boxed{\frac{10}{3}} \right]^2+3\implies f(~~g(2)~~)=\cfrac{10^2}{3^2}+3
\\\\\\
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2 years ago

True or false: you flip the sign every time you multiply or divide when you're solving an

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2 years ago

John is driving around town. When he reaches the gas station, he notes that he has traveled 20 miles. He reaches home 2 hours la

Troyanec [42]

Answer:

$d=5t+20$

Step-by-step explanation:

The equation that will model this situation will be of the form

$d=mt+b$

where $t$ is the time in hours john has traveled since the gas station, and $d$ is the distance.

Now we know that John has already traveled 20 miles when he is at the gas station, this means at $t=0$ , $d=20$ ; or

$20=m(0)+b$

$\boxed{b=20}$

Thus we have

$d=mt+20$ .

Now we need to figure out $m.$

When John reaches home 2 hours later he notes that he has traveled 30 miles, which means he has traveled 30 - 20 = 10 miles; thus we have

$m= \frac{\Delta d}{\Delta t} =\frac{10 miles}{2hours} =5$

$\boxed{m=5}$

Now we have the full equation:

$\boxed{d=5t+20}$

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2 years ago

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which wouldn't be a valid solution over any interval containing $x=0$ .

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