All categories

3 years ago

E bisects line segment df. df=2y, and ef=8y-3. find de, ef, and df.

Mathematics

1 answer:

Usimov [2.4K]3 years ago

8 0

1/2DF = EF

1/2(2y) = 8y - 3

Multiply 1/2 * 2y.

y = 8y - 3

Subtract 8y from both sides.

-7y = -3

Divide both sides by -7.

y = 3/7

Plug 3/7 for y into DF and EF.

DF: 2(3/7) = 6/7

EF: 8(3/7) - 3 = 24/7 - 3 = 3/7

EF = DE because they are bisected, so DE = 3/7

You might be interested in

If f(x) = 3x - 7, then f-1(x)=

fgiga [73]

Answer:

Step-by-step explanation:

so f-1(x) means we are finding the inverse

f(x) = 3x - 7.....change f(x) to y

y = 3x - 7.....now switch ur x and y and solve for y

x = 3y - 7....add 7 to both sides

x + 7 = 3y...divide both sides by 3

(x/3) + 7/3 = y....rearrange

y = (x/3) + 7/3....can also be written as : y = 1/3x + 7

now just change ur y to f-1(x)

f-1(x) = x/3 + 7/3.....or f-1(x) = 1/3x + 7/3

4 0

3 years ago

Read 2 more answers

How do you find pie?

KiRa [710]

Well pie is 3.14 etc.
And the way to find that out is by:
~Using the formula.
~To find out what the circumference is you have to use this formula
C= π*d =2*π*r
~Pi equals a circle's circumference divided by its diameter.
~If you plug in the numbers in a calculator the numbers should be 3.14.

4 0

3 years ago

Read 2 more answers

A marketing firm is considering making up to three new hires. Given its specific needs, the management feels that there is a 50%

IRISSAK [1]

Answer:

a) 0.9

b) Mean = 1.58

Standard Deviation = 0.89

Step-by-step explanation:

We are given the following in the question:

A marketing firm is considering making up to three new hires.

Let X be the variable describing the number of hiring in the company.

Thus, x can take values 0,1 ,2 and 3.

$P(x\geq 2) = 50\%= 0.5\\P(x = 0) = 10\% = 0.1\\P(x = 3) = 18\% = 0.18$

a) P(firm will make at least one hire)

$P(x\geq 2) = P(x=2) + P(x=3)\\0.5 = P(x=2) + 0.18\\ P(x=2) = 0.32$

Also,

$P(x= 0) +P(x= 1) + P(x= 2) + P(x= 3) = 1\\ 0.1 + P(x= 1) + 0.32 + 0.18 = 1\\ P(x= 1) = 1- (0.1+0.32+0.18) = 0.4$

$\text{P(firm will make at least one hire)}\\= P(x\geq 1)\\=P(x=1) + P(x=2) + P(x=3)\\ = 0.4 + 0.32 + 0.18 = 0.9$

b) expected value and the standard deviation of the number of hires.

$E(X) = \displaystyle\sum x_iP(x_i)\\=0(0.1) + 1(0.4) + 2(0.32)+3(0.18) = 1.58$

$E(x^2) = \displaystyle\sum x_i^2P(x_i)\\=0(0.1) + 1(0.4) + 4(0.32) +9(0.18) = 3.3\\V(x) = E(x^2)-[E(x)]^2 = 3.3-(1.58)^2 = 0.80\\\text{Standard Deviation} = \sqrt{V(x)} = \sqrt{0.8036} = 0.89$