I need help plssssssssssssssssssssssss

Which equation represents a parabola that has a focus of (0, 0) and a directrix of y=−6 ? x² = 12y x2=12(y+3) x2=−3(y+3) x2=−3y

Svetlanka [38]

Did you get the answer?

3 0

3 years ago

Read 2 more answers

Tom invests $100,000 at 4.9% interest compounded annually. When will Tom be a millionaire?

Mrac [35]

Millionare means having $1 million or more.
He will be a millionare in 2 years... I think

4 0

4 years ago

A 20 foot ladder is leaning up against the side of a house the distance between the house and the base of the ladder is 4 feet f

Pie

Answer:

78.5°

Step-by-step explanation:

We solve for the above question, using the formula for the Trigonometric function of Cosine

cos θ = Adjacent/Hypotenuse

Adjacent = The distance between the house and the base of the ladder = 4 feet

Hypotenuse = Length of the Ladder = 20 feet

Hence,

cos θ = 4/20

θ = arc cos(4/20)

θ = 78.463040967°

Approximately = 78.5°

Therefore, the angle that the ladder makes with the ground is 78.5°

6 0

3 years ago

Do anyone know how to do this?

kykrilka [37]

Answer:

1.) x=4

2.) x=14

Step-by-step explanation:

FOR #1:

1.) In order to solve the parallelogram, you must know that each opposite sides must equal...

so KL=JM and KJ=LM

2.) To solve for "x", you must substitute

$KL= 7x-2\\JM=12x-22$

so it should end up with...

$7x-2=12x-22$

3.) Solve the equation with some algebra

$7x-2=12x-22\\-5x=-20\\x=4$

FOR #2:

1.) In order for a square, rectangle, parallelogram, (or any shape with four sides), it must equal to 360°. And you just have two given angles, which equal to 180, so that means the unknown equals to 180...So in this case you should have

$(3x+5)+(9x-17)+180=360$

as your equation...

2.) Solve with algebra

$(3x+5)+(9x-17)+180=360$

$3x+5+9x-17=180\\12x-12=180\\(\frac{1}{12}) 12x=168 (\frac{1}{12})\\x=14$

multiplied

3 0

3 years ago

Read 2 more answers

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) Firstly, we will calculate Median for Class A;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) Now, we will calculate Median for Class B;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago