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3 years ago

9

Please help me with these!

1 answer:

katen-ka-za [31]3 years ago

3 0

The correct answer would be D

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Determine the product, h (x), of the linear and quadratic factors.

Wewaii [24]

$h(x) =8x^3-19x^2+14x-3$

Step-by-step explanation:

Given functions are:

$f(x) = 8x-3\\g(x) = (x-1)^2$

h(x) is the product of g(x) and f(x)

$h(x) = f(x) * g(x)\\= (8x-3)(x-1)^2\\= (8x-3)(x^2-2x+1)\\= 8x(x^2-2x+1)-3(x^2-2x+1)\\= 8x^3-16x^2+8x-3x^2+6x-3$

Combining like terms

$=8x^3-16x^2-3x^2+8x+6x-3\\=8x^3-19x^2+14x-3$

Hence,

$h(x) =8x^3-19x^2+14x-3$

Keywords: Functions, product

Learn more about functions at:

#LearnwithBrainly

8 0

3 years ago

How do I calculate the value of x?

S_A_V [24]

Answer:

niewiem.

Step-by-step explanation:

nwieime.

5 0

3 years ago

Read 2 more answers

Two factors of 30 add up to 9 what are they

Readme [11.4K]

The factors of 20 are 1,2,3,5,6,10,15, and 30
6+3=9.

4 0

3 years ago

x axis is x = 4.00 - 7.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) s

andrezito [222]

Answer:

(a) 0 s

(b) 4.00 m

(c) -0.76 s

(d) +0.76 s

Step-by-step explanation:

$x=4.00-7.00t^2$

It stops momentarily when the velocity, $v$ is 0. $v$ is the derivate of $x$ .

(a) $v=\frac{dx}{dt}=-14.00t$

Setting this to 0,

$-14.00t=0$

$t=0$

(b) Substitute this value for $t$ in $x$ to get its position.

$x=4.00-7.00\times0^2=4.00$ m

It passes the origin when $x=0$

$4.00-7.00t^2=0$

$7.00t^2=4$

$t^2=\frac{4}{7}$

$t=\pm\sqrt{\frac{4}{7}}$

(c) The negative time is $t=-\sqrt{\frac{4}{7}} =-0.76 s$

(d) The positive time is $t=+\sqrt{\frac{4}{7}} =+0.76 s$

8 0

3 years ago

"What is the value of x in the equation, considering the circle?"

AveGali [126]

Can I see an image ?
I cannot answer without seeing the circle

5 0

3 years ago