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3 years ago

Using the vertical line test. Is this a function?

Mathematics

1 answer:

marta [7]3 years ago

7 0

Answer:

Can't get one(Read explanation)

Step-by-step explanation:

What you do to find out if its a function is see if any of the dots are in the same x value. If they are, its not a function. If they aren't it is a function.

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Mei Mei invested $1700 in an account that pays 3% interest compounded annually.

algol [13]

Answer:

$2728

Step-by-step explanation:

Use compound interest formula:

$A = P(1 + \frac{r}{n})^{nt}$

P = 1700, r = 0.03, n = 1, t = 16

Plug in numbers:

$A = 1700(1 + \frac{0.03}{1})^{(1)(16)}$

Simplify:

$A = 1700(1.3)^{16$

Use calculator:

$A = 2728$

4 0

2 years ago

2) Phillip has 8 red balls, 3 green balls, 6 yellow balls, 3 orange balls, 13 black balls

AVprozaik [17]

Mean 6.6
Median 6
Mode 3
Range 10

not positive on the 6.6 but the rest should be good

5 0

3 years ago

Derive the equation of the parabola with a focus at (0, −4) and a directrix of y = 4.

CaHeK987 [17]

<u>Answer:</u>

$y=-\frac{1}{16} x^2$

<u>Step-by-step explanation:</u>

Equation of parabola with focus at (0,-4) and directrix is $y=4$ .

We know that parabola is the locus of all the points such that the distance from fixed point on the parabola to fixed line directrix is the same.

The parabola is opening downwards.

Let any point on parabola is (x,y).

Distance from focus(0,-4) to (x,y) = $\sqrt{(x-0)^2+(y-(-4))^2}=|y-4|$

$\sqrt{(x-0)^2+(y-+4)^2}=|y-4|$

$(x-0)^2+(y+4)^2=(y-4)^2$

$(x)^2+(y+4)^2=(y-4)^2$

$x^2+(y^2+4y+4y+16)=y^2-4y-4y+16$

$x^2+y^2+8y+16=y^2-8y+16$

$x^2+16y^2=0$

$16y=-x^2$

$y=-\frac{1}{16} x^2$

6 0

2 years ago

On the coordinate plane below, triangle ABC was rotated

agasfer [191]

Answer: AB/y = BC/z = AC/x

Step-by-step explanation:

Side AB corresponds to side y, side BC corresponds to side z, and side AC corresponds to side x.

7 0

2 years ago

What is the cube root of -216?

Tatiana [17]

(-6)*(-6)*(-6)=-216.

Therefore, the cubed root of -216 is -6.

4 0

3 years ago