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3 years ago

I'll give brainliest

Mathematics

2 answers:

avanturin [10]3 years ago

7 0

Hey There!

In order to solve this you need to factor f(x) = x^3 - 19x + 30 Entirely.

When you do that you get these factors;

f(x) = (x – 2) (x – 3) (x + 5)

From there you play 'elimination'

A) There is no listed factor for (x + 2)

B) There is no listed factor for (x + 3)

C) There is a listed factor for (x + 5)!

D) There is no listed factor for (x - 5)

Hope I helped, Five Stars & Brainliest always helps!

erma4kov [3.2K]3 years ago

6 0

Answer:

I think its D sorry if I get it wrong

Step-by-step explanation:

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Help me quick gotta get a 100% before 12

Arada [10]

Answer:

Step-by-step explanation:

If it's a 15 dollar variable fee and a flat 10 our equation looks like this

15c+10

which is the same as A

Now that we know the equation we can just plug a couple of numbers in

when c=3 what is h

15(3)+10=

THerefore the answer is B

8 0

3 years ago

Read 2 more answers

We want to find the zeros of this polynomial:

slava [35]

Answer:

x=-3 x=-1 x=2 x=-2

Step-by-step explanation:

p(x) = (x^2 + 4x + 3)(x^2 – 4)

Set this equal to zero to find the x intercepts

0 = (x^2 + 4x + 3)(x^2 – 4)

Using the zero product property

(x^2 + 4x + 3) =0 (x^2 – 4) =0

Factor

(x+3)(x+1) =0 (x-2) (x+2)=0

Using the zero product property

x+3 =0 x+1 =0 x-2 =0 x+2 =0

x=-3 x=-1 x=2 x=-2

7 0

3 years ago

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2

Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

$y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\$

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t = e^−2(0) = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

$\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt} = -2te^{-2t} = -2(0)e^{-2(0)} = 0$

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0

3 years ago

HELP ASAP!!!!!!! I NEED THIS SOON!!!

Arte-miy333 [17]

Answer:

Is this a question about the laws of angle of incidence or is it just coordinates

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A triangle has angle measures of 47 degrees and 92 degrees. What is the measure of the third angle?

ludmilkaskok [199]

Answer:

The missing angle is 41 degrees

Step-by-step explanation:

First of all, a triangle is measured at 180 degrees in total

To find the missing measurement for the third angle, add the 47 and 92 degrees together which would equal 139 degrees.

47+92=139 degrees

Next just subtract 139 from 180 to get your answer.

180-139=41 degrees

The answer is 41 degrees

Hope this helps

5 0

3 years ago