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3 years ago

Help please!!!!help help help

Mathematics

2 answers:

Iteru [2.4K]3 years ago

7 0

Answer:

{-2,-1, 0, 1, 8}

Step-by-step explanation:

Domain is the input values ie. x values

{-2,-1, 0, 1, 8}

OverLord2011 [107]3 years ago

5 0

Answer:

ask the goggle not 1 you help in thi aap bro your aand me same thing make sorry

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HEY!! I’M REALLY DESPERATE RN FOR AN ANSWER SO LIKE HERE’S ALL MY POINTS!! Please show the work!

BabaBlast [244]

Answer:

see explanation

Step-by-step explanation:

One way to divide is to use the divisor as a factor in the numerator

Consider the numerator

x²(x - 3) + 3x² - x + 6

= x²(x - 3) + 3x(x - 3) + 9x - x + 6

= x²(x - 3) + 3x(x - 3) + 8(x - 3) + 24 + 6

= x²(x - 3) + 3x(x - 3) + 8(x - 3) + 30

quotient = x² + 3x + 8 , remainder = 30

$\frac{x^3-x+6}{x-3}$ = x² + 3x + 8 + $\frac{30}{x-3}$

6 0

3 years ago

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If the standard deviation of a data set was originally 8, and if each value in the data set was multiplied by 8.5, what would be

maxonik [38]

Answer:

D. 68 (APEX).........................................................................................

6 0

3 years ago

The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing (in mm3/s) when the diameter is 40 m

daser333 [38]

Answer:

$\dfrac{dV}{dt}=502.65\ mm^3/s$

Step-by-step explanation:

The volume of a sphere is given by :

$V=\dfrac{4}{3}\pi r^3$

The rate of change of volume means,

$\dfrac{dV}{dt}=\dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt}=\dfrac{4}{3}\pi \times 3r\times \dfrac{dr}{dt}$

We have, $\dfrac{dr}{dt}=2\ mm/s\ and\ r=40\ mm$

So,

$\dfrac{dV}{dt}=\dfrac{4}{3}\pi \times 3\times 20\times 2\\\\=502.65\ mm^3/s$

So, the volume is increasing at the rate of $502.65\ mm^3/s$ .

6 0

3 years ago

Fill in the blank.. HELPPP PLEASEE!!!!!!

Nadya [2.5K]

Answer:

real number

cannot, integers

Step-by-step explanation:

4 0

3 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers