Is this a function hmm

Y = 3/2x + 15, provide the x-y coordinates of the y-intercept and the x-intercept. Show your work.

Ray Of Light [21]

Answer:

The y-intercept is 15, which means the ordered pair is (0, 15), and the x-intercept is -10, which means the ordered pair is (-10, 0)

Step-by-step explanation:

To find the y-intercept it is quite easy. Since this is written in slope-intercept form, we can see the y intercept as the constant at the end of the equation.

The x-intercept can be found by inputting 0 for y and then solving for x.

y = 3/2x + 15

0 = 3/2x + 15

-15 = 3/2x

-10 = x

5 0

3 years ago

This hanger is in balance. There are two labeled weights of 4 grams and 12 grams. The three circles each have the same weight.

antoniya [11.8K]

Answer:

8/3

Step-by-step explanation:

6 0

3 years ago

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30 more than the quotient of b and c

Aliun [14]

Answer:

b/c+30

Step-by-step explanation:

8 0

2 years ago

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Find the nth term of this number sequence<br>15, 12, 9, 6....

Ann [662]

Answer:

You only have four terms so n can only be 1, 2, 3 or 4!

I think you mean: What is the nth term of the sequence 15, 12, 9, 6, …?

The ellipsis (…) tells us that the sequence continues.

This is an arithmetic sequence, where the general term is: a+bn

n=1:a+b=15 [Eq. 1]

n=2:a+2b=12 [Eq. 2]

Subtracting Eq. 1 from Eq.2: b=−3

Substituting this into Eq. 1: a−3=15⇒a=18

Thus the nth term is: 18−3n

Step-by-step explanation:

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8 0

3 years ago

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v

olga_2 [115]

We are given

$f=z^3 -x^2y$

Firstly, we can find gradient

so, we will find partial derivatives

$f_x=0 -2xy$

$f_x=-2xy$

$f_y=0 -x^2$

$f_y=-x^2$

$f_z=3z^2$

now, we can plug point (-5,5,2)

$f_x=-2*-5*5=50$

$f_y=-(-5)^2=-25$

$f_z=3(2)^2=12$

so, gradient will be

$gradf=(50,-25,12)$

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

$|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}$

$|v|=\sqrt{29}$

now, we can find unit vector

$v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

now, we can find dot product to find direction of the vector

$dir=(gradf) \cdot (v')$

now, we can plug values

$dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

$dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })$

$dir=-\frac{248\sqrt{29}}{29}$ .............Answer

7 0

4 years ago

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