All categories

1 year ago

3x^2-39+120 factor this

Mathematics

2 answers:

mylen [45]1 year ago

6 0

Answer:

$9x^{2} +81$

Step-by-step explanation:

I Guess.

Irina18 [472]1 year ago

6 0

Answer:

3(x^2 + 27)

My answer was totally off! I just re evaluated the equation and I got this:

LMK if it's wrong :)

You might be interested in

Solve for n. 2/3(1+n)=-1/2n

notka56 [123]

Answer:

=−4/7

Step-by-step explanation:

And yh I'm sorry if I got it wrong

5 0

3 years ago

Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the paraboloid z = 9x2 y2 and the p

dolphi86 [110]

The vector function is, r(t) = $\bold{ < t,2t^2,9t^2+4t^4 > }$

Given two surfaces for which the vector function corresponding to the intersection of the two need to be found.

First surface is the paraboloid, $z=9x^2+y^2$

Second equation is of the parabolic cylinder, $y=2x^2$

Now to find the intersection of these surfaces, we change these equations into its parametrical representations.

Let x = t

Then, from the equation of parabolic cylinder, $y=2t^2$ .

Now substituting x and y into the equation of the paraboloid, we get,

$z=9t^2+(2t^2)^2 = 9t^2+4t^4$

Now the vector function, r(t) = <x, y, z>

So r(t) = $\bold{ < t,2t^2,9t^2+4t^4 > }$

Learn more about vector functions at brainly.com/question/28479805

#SPJ4

7 0

1 year ago

A car is driving down I-55 approaching the exit for two cities: Angleton and Bisectorville. Both towns are 3.14 miles to I-55 an

Shtirlitz [24]

Answer:

3.14 mile-55=51.86 miles from the car and exit

i think i help.(sorry if it is not correct)

Step-by-step explanation:

7 0

3 years ago

Hudson was working on the problem 1/8y=4 He thought he should multiply bolth sides of the equation by 8 to get 32.

kaheart [24]

Answer: Use the given functions to set up and simplify

32. 1/8y = 4He= y =32He8= y =32He32=y=32He

7 0

2 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

3 years ago