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3 years ago

WILL GIVE BRAINLIEST

Mathematics

1 answer:

sdas [7]3 years ago

5 0

3. f(-6) = 12+1 =13

f(-2) = 4+1 = 5

f(0) =1

Range {1,5,13}

4. f(-2) = (-2)^3+1 =-7

f(-1) = (-1)^2 +1 =0

f(3) = (3)^3 +1 = 28

Range = {-7,0,28}

5.the sequence is arithmetic

d= -11+19 = 8

an = a1 + d(n-1)

an = -19 +8(n-1)

6.l =w+5

a =l*w

a(w) =(w+5) * w

a(w)= w^2 +5w

f(w) = w^2 +5w

f(8) = 8^2 +5(8)

f(8) = 64 +40

f(8) =104 in^2

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Find all solutions to x³+3x² − 9x − 27 = 0 by factoring the equation.

SCORPION-xisa [38]

Answer:

The solutions are $x=-3,\:x=3$

Step-by-step explanation:

To factor this cubic polynomial $x^3+3x^2-9x-27$ you must:

Group the polynomial into two sections

$x^3+3x^2-9x-27=\left(x^3+3x^2\right)+\left(-9x-27\right)$

Factor out -9 from $(-9x-27)$

$(-9x-27)=-9\left(x+3\right)$

Factor out $x^2$ from $(x^3+3x^2)$

$(x^3+3x^2)=x^2\left(x+3\right)$

$x^3+3x^2-9x-27=-9\left(x+3\right)+x^2\left(x+3\right)$

Factor out common term $x+3$

$-9\left(x+3\right)+x^2\left(x+3\right)=\left(x+3\right)\left(x^2-9\right)$

$x^3+3x^2-9x-27=\left(x+3\right)\left(x^2-9\right)$

Factor $x^2-9$

$x^2-9=\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27= \left(x+3\right)\left(x+3\right)\left(x-3\right)$

$x^3+3x^2-9x-27=\left(x+3\right)^2\left(x-3\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+3=0, \quad{x=-3}\\x-3=0, \quad{x=3}$

The solutions are

$x=-3,\:x=3$

3 0

3 years ago

NEED HELP ASAP PLEASE

AnnZ [28]

Answer:

8.5%

Step-by-step explanation:

I=PRT

85= 2000*r*0.5

85=1000r

0.085=r

8.5%

6 0

3 years ago

PLEASE HELP ME QUICK!!!Can you form a triangle with angle measurements of 50°, 50° and 100° explain why or why not.

alex41 [277]

Answer:

No, you can't form a triangle because angles in a triangle add up to 180 degrees.

Step-by-step explanation:

100+50+50= 200 degrees, which means that you cant form a triangle with these measurements because angles in a triangle have to add up to 180 degrees.

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Furkat [3]

Answer:

Step-by-step explanation:

It would be 40

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3 years ago

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andrew11 [14]

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3 years ago