All categories

3 years ago

Which graph indicates decreasing speed? A. Graph c B Graph A c graph F d graph

Mathematics

1 answer:

Bas_tet [7]3 years ago

6 0

I think that it would be option A. Graph C. It shows the line going down from the y-axis to the x-axis, so it would be graph C.

You might be interested in

Help I’ll mark you brainly!!

Ulleksa [173]

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of non-zero numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

6 0

3 years ago

Find the unit rate. 16 miles in 4 hours

MatroZZZ [7]

Answer:

4 miles per hour

Step-by-step explanation:

divide 16 by 4

7 0

4 years ago

Can you please help on this?

krek1111 [17]

Answer:

7.6×7.6= 57.76

so it has to be bigger than √55

gud luck

7 0

3 years ago

What is the following difference?<br> 2ab2/192an? -5 V81a4 b5

diamong [38]

Answer:

Step-by-step explanation:

2ab * sqrt[3](192ab^2) - 5 * sqrt[3](81 a^4b^5) =

= 2ab * sqrt[3](4^3 * 3 * a * b^2) - 5 * sqrt[3](3^3 * 3 * a^3 * a * b^3 * b^2)

= 2ab * 4 sqrt[3](3 * a * b^2) - 5 * 3 * a * b * sqrt[3](3 * a * b^2)

= 8ab * sqrt[3](3 * a * b^2) - 15ab * sqrt[3](3 * a * b^2)

= -7ab * sqrt[3](3ab^2)

Answer: C

7 0

3 years ago

Prove that a cubic equation x 3 + ax 2 + bx+ c = 0 has 3 roots by finding the roots.

evablogger [386]

That's a pretty tall order for Brainly homework. Let's start with the depressed cubic, which is simpler.

Solve

$y^3 + 3py = 2q$

We'll put coefficients on the coefficients to avoid fractions down the road.

The key idea is called a split, which let's us turn the cubic equation in to a quadratic. We split unknown y into two pieces:

$y = s + t$

Substituting,

$(s+t)^3 + 3p(s+t) = 2q$

Expanding it out,

$s^3+3 s^2 t + 3 s t^2 + t^3 + 3p(s+t) = 2q$

$s^3+t^3 + 3 s t(s+t) + 3p(s+t) = 2q$

$s^3+t^3 + 3( s t + p)(s+t) = 2q$

There a few moves we could make from here. The easiest is probably to try to solve the simultaneous equations:

$s^3+t^3=2q, \qquad st+p=0$

which would give us a solution to the cubic.

$p=-st$

$t = -\dfrac p s$

Substituting,

$s^3 - \dfrac{p^3}{s^3} = 2q$

$(s^3)^2 - 2 q s^3 - p^3 = 0$

By the quadratic formula (note the shortcut from the even linear term):

$s^3 = q \pm \sqrt{p^3 + q^2}$

By the symmetry of the problem (we can interchange s and t without changing anything) when s is one solution t is the other:

$s^3 = q + \sqrt{p^3+q^2}$

$t^3 = q - \sqrt{p^3+q^2}$

We've arrived at the solution for the depressed cubic:

$y = s+t = \sqrt[3]{q + \sqrt{p^3+q^2}} + \sqrt[3]{ q - \sqrt{p^3+q^2} }$

This is all three roots of the equation, given by the three cube roots (at least two complex), say for the left radical. The two cubes aren't really independent, we need their product to be $-p=st$ .

That's the three roots of the depressed cubic; let's solve the general cubic by reducing it to the depressed cubic.

$x^3 + ax^2 + bx + c=0$

We want to eliminate the squared term. If substitute x = y + k we'll get a 3ky² from the cubic term and ay² from the squared term; we want these to cancel so 3k=-a.

Substitute x = y - a/3

$(y - a/3)^3 + a(y - a/3)^2 + b(y - a/3) + c = 0$

$y^3 - ay^2 + a^2/3 y - a^3/27 + ay^2-2a^2y/3 + a^3/9 + by - ab/3 + c =0$

$y^3 + (b - a^2/3) y = -(2a^3+9ab) /27$

Comparing that to

$y^3 + 3py = 2q$

we have $p = (3b - a^2) /9, q =-(a^3+9ab)/54$

which we can substitute in to the depressed cubic solution and subtract a/3 to get the three roots. I won't write that out; it's a little ugly.

8 0

4 years ago