All categories

2 years ago

PLEASE HELP, WILL GIVE BRAINLIEST + 100 POINTS

Mathematics

1 answer:

attashe74 [19]2 years ago

3 0

Answer:

part A.equation of an ellipse

part B.equation of a hyperbola

part C.domain of each conic section

You might be interested in

You and your best friend decide to go to dinner at Chickfila. You each

svetoff [14.1K]

Answer:

= $28.90

Step-by-step explanation:

8 0

2 years ago

Keisha says that all functions are relations but not all relations are functions. kevin says that all relations are functions bu

vladimir1956 [14]

Keisha is correct, because as per the definition A function is a special relationship where each input has a single output.

A function is a special relation. In other words, a relation if and only if it has a specific characteristic where each input has a single output, then it is called a Function.

All functions are relations but not all relations are functions.

3 0

3 years ago

Write the equation for a parabola with a focus at (6,-4) and a directrix at y= -7

shtirl [24]

Given:

The focus of the parabola is at (6,-4).

Directrix at y=-7.

To find:

The equation of the parabola.

Solution:

The general equation of a parabola is:

$y=\dfrac{1}{4p}(x-h)^2+k$ ...(i)

Where, (h,k) is vertex, (h,k+p) is the focus and y=k-p is the directrix.

The focus of the parabola is at (6,-4).

$(h,k+p)=(6,-4)$

On comparing both sides, we get

$h=6$

$k+p=-4$ ...(ii)

Directrix at y=-7. So,

$k-p=-7$ ...(iii)

Adding (ii) and (iii), we get

$2k=-11$

$k=\dfrac{-11}{2}$

$k=-5.5$

Putting $k=-5.5$ in (ii), we get

$-5.5+p=-4$

$p=-4+5.5$

$p=1.5$

Putting $h=6, k=-5.5,p=1.5$ in (i), we get

$y=\dfrac{1}{4(1.5)}(x-6)^2+(-5.5)$

$y=\dfrac{1}{6}(x-6)^2-5.5$

Therefore, the equation of the parabola is $y=\dfrac{1}{6}(x-6)^2-5.5$ .

4 0

2 years ago

Suppose the lengths of two sides of a right triangle are represented by 2x and 3 (x + 1), and the longest side is 17 units. Find

densk [106]

Answer:

x=4

Step-by-step explanation:

Step 1:-

given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.

AB = 2x and BC = 3(x+1) and longest side AC= 17

by using Pythagoras theorem

$AC^2 = AB^2 + BC^2$

step 2:-

The hypotenuse is longest side is AC = 17 units

(17)^2 = 4x^2 +9(X+1)^2

on simplification, we will use formula

$(a + b)^2 = a^2 +2ab+b^2$

289 = 4x^2 +9(x^2+2x+1)

$13x^2 +18x-280 = 0$

finding factors 70 X 52 = 3640

$13x^2 +70x-52x-280 = 0$

$13x^2 -52x+ 70x-280 = 0$

Taking common , we get

13x(x-4)+70(x-4)=0

x-4=0 and 13x+70=0

x=4 and $13x =-70$

x=4 and $x=\frac{-70}{13}$

we can not choose negative value so x value is 4

Final answer:- x = 4

verification:-

$AC^2 = AB^2 + BC^2$

289 = 4(4)^2+9(4+1)^2

289 = 64 +9(25)

289=289

6 0

3 years ago

Read 2 more answers

Need help

aleksandr82 [10.1K]

Using the normal distribution, the probabilities are given as follows:

a. 0.4602 = 46.02%.

b. 0.281 = 28.1%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

The parameters are given as follows:

$\mu = 959, \sigma = 263, n = 37, s = \frac{263}{\sqrt{37}} = 43.24$

Item a:

The probability is one subtracted by the p-value of Z when X = 984, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{984 - 959}{263}$

Z = 0.1

Z = 0.1 has a p-value of 0.5398.

1 - 0.5398 = 0.4602.

Item b:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{984 - 959}{43.24}$

Z = 0.58

Z = 0.58 has a p-value of 0.7190.

1 - 0.719 = 0.281.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0

2 years ago