All categories

3 years ago

If f(z)== +2, what is f(4)?

Mathematics

1 answer:

Bad White [126]3 years ago

3 0

Answer:

soudgqa

Step-by-step explanation:

You might be interested in

The length of a rectangle is twice its width. If the perimeter of the rectangle is 54 in, find its area.

sergiy2304 [10]

Answer: 162 in²

Step-by-step explanation:

The factors of 54 are listed below:

18 and 9 are the two factors that when plugged into the equation below, equals 54:

$l+l+w+w=P$

This can also be written as:

$(l*2)+(w*2)=P$

So now that we have our two factors, we need to multiply them by each other to get our answer.

18 times 9 = 162

Our answer is 162

6 0

3 years ago

Read 2 more answers

Solve everything for brainiest

IRISSAK [1]

Answer:

1. $a = 8$

2. $b = \frac{1}{8}$

3. $c = 2 \frac{3}{5}$

4. $d = \frac{13}{50}$

Step-by-step explanation:

6 0

3 years ago

H/5 = 5.1<br> What is H?

inysia [295]

This basically says all the steps ! If you have any questions just tell me :)

5 0

3 years ago

Read 2 more answers

Is it possible to draw an acute isosceles triangle with side lengths of 6 cm, 9 cm, and 12 cm and angles of 30°, 50°, and 100° ?

bagirrra123 [75]

Answer:

Yes

Step-by-step explanation:

A full triangle is 180 degrees. 100 + 30 + 50 = 180.

I hope you get it right! :D

(i hope i get brainlist)

5 0

3 years ago

(5x^3-8x^2+9x+12)/(x-3)<br><br> please answer with sentences on how to do it step by step. thanks!

Anastaziya [24]

Answer:

$5x^2+7x+30+\frac{102}{x-3}$

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}$

6 0

3 years ago

Read 2 more answers