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2 years ago

What is the product of 8x-3 and x2- 4x +8

1 answer:

5 0

A product is the answer that you get when you multiply numbers together. So for this problem, you have 2 groups to multiply together. Since I cannot show a square or cubed x, I will put an x2 for x squared and an x3 for x cubed. You have to multiply each number in the first parentheses by each number in the second parentheses. Then combine any like sets.

(8x-3)(x2-4x+8)
8x3-32x2+64x-6x+12x-24
8x3-32x2+70x-24

So the answer is 8x cubed minus 32x squared plus 70x minus 24. Whew! That's a long one. Hope I didn't miss anything.

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WILL GIVE BRANLIEST!!! Pls help! Determine the coordinates of the point on the straight line y=3x+1 that is equidistant from the

iren [92.7K]

Let , coordinate of points are P( h,k ).

Also , k = 3h + 1

Distance of P from origin :

$d=\sqrt{h^2+k^2}$

Distance of P from ( -3, 4 ) :

$d=\sqrt{(h+3)^2+(k-4)^2}$

Now , these distance are equal :

$h^2+(3h+1)^2=(h+3)^2+(3h+1-4)^2\\\\h^2+(3h+1)^2=(h+3)^2+(3h-3)^2$

Solving above equation , we get :

$P=(\dfrac{16}{21},\dfrac{23}{7})$

Hence , this is the required solution.

6 0

2 years ago

Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right

Jet001 [13]

Answer:

Option B.

Step-by-step explanation:

It is given that ΔSRQ is a right angle triangle, ∠SRQ is right angle.

RT is altitude on side SQ, ST=9, TQ=16 and SR=x.

In ΔSRQ and ΔSTR,

$m\angle S=m\angle S$ (Reflexive property)

$m\angle R=m\angle T$ (Right angle)

By AA property of similarity,

$\triangle SRQ\sim \triangle STR$

Corresponding parts of similar triangles are proportional.

$\dfrac{SR}{SQ}=\dfrac{ST}{SR}$

Substitute the given values.

$\dfrac{x}{9+16}=\dfrac{9}{x}$

$\dfrac{x}{25}=\dfrac{9}{x}$

On cross multiplication we get

$x^2=25\times 9$

$x^2=225$

Taking square root on both sides.

$x=\sqrt{225}$

$x=15$

The value of x is 15. Therefore, the correct option is B.

7 0

2 years ago

Read 2 more answers

A farmer kept 96 kilograms of animal feed in his barn.

lesya692 [45]

Answer:

The amount of animal feed left was 86.4 kilograms.

Step-by-step explanation:

Initial amount of animal feed in the barn was 96 kilograms.

The barn leaked and $\frac{1}{10}$ of the feed was wasted.

So, the amount of feed wasted $=\frac{1}{10}(96)=9.6$ kilograms.

Now, for finding the amount of feed that was left, we need to subtract the amount of wasted feed from the initial amount of feed.

So, the amount of animal feed left $=(96-9.6)\ kilograms=86.4\ kilograms$

4 0

2 years ago

Read 2 more answers

On a number line, the coordinates of X, Y, Z, and W are , , , and , respectively. Find the lengths of the two segments be

Arturiano [62]

Complete Question:

On a number line, the coordinates of X, Y, Z, and W are −8, −5, 4, and 6, respectively. Find the lengths of the two segments below. Then tell whether they are congruent. $\overline{XY}$ and $\overline{ZW}$

Answer:

$\overline{XY} = 3$

$\overline{ZW} = 2$

They are not congruent

Step-by-step explanation:

Length of segment XY:

Coordinate of X = -8

Coordinate of Y = -5

$\overline{XY}$ = |-8 -(-5)| = |-8 + 5| = 3

Length of ZW:

Coordinate of Z = 4

Coordinate of W = 6

$\overline{ZW}$ = |4 - 6| = 2

$\overline{XY}$ ≠ $\overline{ZW}$ , therefore, they are not congruent.

3 0

2 years ago

Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words

Taya2010 [7]

Answer:

Step-by-step explanation:

Answer:

a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)

Explanation:

To solve these equations the method of direct integration is the easiest.

a) the given equation is

dy / dt = and -8

dy / y-8 = dt

We change variables

y-8 = u

dy = du

We replace and integrate

∫ du / u = ∫ dt

Ln (y-8) = t

We evaluate at the lower limits t = 0 for y = y₀

ln (y-8) - ln (y₀-8) = t-0

Let's simplify the equation

ln (y-8 / y₀-8) = t

y-8 / y₀-8 =

y-8 = (y₀-8)

b) the equation is

dy / dt = 2y -5

u = 2y -5

du = 2 dy

du / 2u = dt

We integrate

½ Ln (2y-5) = t

We evaluate at the limits

½ [ln (2y-5) - ln (2y₀-5)] = t

Ln (2y-5 / 2y₀-5) = 2t

2y -5 = (2y₀-5)

c) the equation is very similar to the previous one

u = 2y -10

du = 2 dy

∫ du / 2u = dt

ln (2y-10) = 2t

We evaluate

ln (2y-10) –ln (2y₀-10) = 2t

2y-10 = (2y₀-10)

4 0

2 years ago