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3 years ago

12

What goes outside during division with fractions? The numerator, or the denominator?

1 answer:

Ymorist [56]3 years ago

6 0

Answer:

Denominator

Step-by-step explanation:

Think of it this way - numerator/denominator = nana/dog

Nana goes inside to the bathroom, but the dog goes outside to do its thing

If it was the other way around, it wouldn't be very pleasant.

LOL

Anyway, I hope this was helpful

Mark as Brainliest if it was

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5,250 as scientific notation

earnstyle [38]

$\text{Hey there!}$

$\text{5.250 as a scientific notation}$

$\bf{The\ answer\ is:}$ $\boxed{\bf{5.25 \times10^3}}}}$

$\text{Because, we had to move your decimal approximately 3 placecs to the left}$

$\rightarrow\tex{= 5250. }$

$\boxed{\text{That's how we know the answer is: \bf{\boxed{5.25 \times10^3}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFirst:)}$

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A geometric figure is a set of a finite number of points

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Can you buttress your point?

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What is the equation of the line?

prohojiy [21]

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A) y = -1/2x + 1/2

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Find the y-intercept(when x = 0), which is 1/2.

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The price of a pair of shoes is 45.90. the sales tax is 5%. how much sales tax do you need to pay

grin007 [14]

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Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

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3 years ago