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3 years ago

For homework in math my problem is g-5=9

Mathematics

2 answers:

sergeinik [125]3 years ago

6 0

G is equal to fourteen (14)

Jobisdone [24]3 years ago

4 0

You didn't say what you're supposed to do with that equation.

Do you need to find out what number 'g' is ?

To do that, just add 5 to each side of the equation.

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Number 17 please help 10 points no links pleAse

larisa86 [58]

Answer:

Step-by-step explanation:

Florida is on the 30 mark so its 30 then Alaska is in between 10 and 20 half so half of these numbers is 15 then your going to subtract 30-15 leaving you with 15 and as you can see that answer is presented

7 0

3 years ago

Help appreciated please and thank you

meriva

Answer:

D. $\frac{4 \sqrt{33} }{33}$

Step-by-step explanation:

$\tan \alpha = \frac{4}{ \sqrt{33} } \\ = \frac{4}{ \sqrt{33} } \times \frac{ \sqrt{33} }{ \sqrt{33} } \\ \\ = \frac{4 \sqrt{33} }{ {( \sqrt{33)} }^{2} } \\ \\ \tan \alpha = \frac{4 \sqrt{33} }{33}$

8 0

3 years ago

Read 2 more answers

Divide 16x3 – 12x2 + 20x – 3 by 4x + 5.

nalin [4]

Answer:

$4x^2 - 8x + 15 - \frac{78}{4x+5}$

Step-by-step explanation:

<em>To solve polynomial long division problems like these, it's helpful to build a long division table. Getting used to building these can make problems like this much simpler to solve.</em>

Begin by looking at the first term of the cubic polynomial.

What would we have to multiply $4x + 5$ by to get an expression containing $16x^3$ ? The answer is $4x^2$ , since $(4x + 5) \times 4x^2 = 16x^2 + 20x$ .

This is the first step of our long division, and we write out the start of our long division table like this:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,16x^3 + 20x^2\\$

On the left is the divisor. On top is $4x^2$ . In the middle is the polynomial we are dividing, and on the bottom is the result of multiplying our divisor by

The next step is to subtract the bottom expression from the middle one, like so:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,0x^3 - 32x^2\\$

We are left with $-32x^2$ . The next thing to do is to add the next term of the polynomial we are dividing to the bottom line, like this:

Now we return to the beginning of the instructions, and repeat the process: namely, what would we have to multiply $4x + 5$ by to get an expression containing $-32x^2$ ? The answer is $-8x$ , and we fill out our long division table like so:

${ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,4x^2 - \,\,\,\,8x\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 - 40x\\$

Once again, we subtract the bottom expression from the one above it, and include the next term of the divisor, like so:

${ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+ 15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\$

And repeat. What do we multiply $4x + 5$ by to get an expression containing $60x$ ? The answer is 15. Our completed long division table looks like this: ${ }\qquad{ }\qquad{ }\quad{ }4x^2 - \,\,\,\,8x \,+15\\4x + 5\quad)\!\!\overline{\,\,\,16x^3 - 12x^2 + 20x - 3}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\underline{16x^3 + 20x^2}\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 32x^2 + 20x\\{ }\qquad{ }\quad{ }\quad{ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{- 32x^2 - 40x}\\{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\qquad{ }\,\,\,\,\,60x - 3\\{ }\qquad{ }\hspace{3cm}\,\,\underline{60x + 75}\\{ }\hspace{4.3cm}\,\,-78$