Ask question

Ask question

All categories

SVETLANKA909090 [29]

1 year ago

10

Simplify the expression.Can you help me with this one. I'm stuck at 12^x4 - 15x^2

1 answer:

Blababa [14]1 year ago

3 0

Simplify the given expression as shown below

$\begin{gathered} 3x(4x^4-5x)=3x*4x^4+3x(-5x)=12x^5+(-15x^2) \\ =12x^5-15x^2 \end{gathered}$ <h2>Therefore, the answer is 12x^5-15x^2</h2>

You might be interested in

How to solve for x and steps on how to do that with the final answer

Maurinko [17]

25/10= 2.5

So, we know that the larger triangle is bigger than the smaller one by a dilation of 2.5.

14*2.5= 35

x= 35

I hope this helps!
~kaikers

6 0

2 years ago

Scoring a hole-in-one is the greatest shot a golfer can make. Once 5 professional golfers each made holes-in-one on the 7th hole

adell [148]

A) Probability that at least one of these golfers makes a hole in one on the 10th hole at the same tournament is

<span>=1−(<span>2493/2494</span><span>)^<span>5
=1-(0.99)^5
= 1- 0.96
=0.4

b) T</span></span></span><span>he probability that all of these golfers make a hole-in-one on the 10th hole at the same tournament
=</span>(<span>1/2494</span><span>)<span>5
</span></span>
which is nearly zero.

5 0

3 years ago

#7:Write the expression in simplest<br> radical form.<br> V24x5y2

vazorg [7]

2x^2y√6x

the image has the answer in a mor clear form

8 0

2 years ago

Phoebe compared the cost of books in 9 different stores.

krek1111 [17]

You have to put the numbers in order and find the middle value so in this case it's-

2, 4, 4, 5, 6, 7, 8, 8, 8

The answer is $6

7 0

3 years ago

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equa

Anvisha [2.4K]

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) : $x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With $\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)

7 0

3 years ago