Witch tipe of line symmetry does the figure have C

D) xºy - 64y4<br>please solve this question<br>First one brainliest

nexus9112 [7]

plz mark me plz plz plz plz plz plz mark me

7 0

3 years ago

PLEASE HELP

alexgriva [62]

Answer:

Step-by-step explanation:

90.25$

5 0

1 year ago

A merchant has coffee worth $20 a pound that she wishes to mix with 70 pounds of coffee worth $90 a pound to get a mixture that

Dvinal [7]

Let us start with assuming the amount of coffee worth $20 a pound to be "x" pounds.

Now she wants to mix "x" pounds of $20 coffee with 70 pounds of $90 coffee.

So the mixture would be (x + 70) pounds.

And the value of mixture would be = 20·x + (90)·(70) = (20x + 6300) dollars.

She want to sell this mixture at rate of $30 a pound. Her earning would be = 30·(x + 70) = (30x + 2100) dollars.

We know that the value of mixture would be equal to her earnings.

30x + 2100 = 20x + 6300

30x - 20x = 6300 - 2100

10x = 4200

x = 420

So, 420 pounds of $20 coffee would be used.

7 0

3 years ago

Read 2 more answers

Herman Melville had worked at the University Medical Center for 20 years when he became permanently disabled while at work. He c

lidiya [134]

Answer:

Herman Melville's annual disability benefit is $46,114.8

Step-by-step explanation:

Given the data in the question;

Herman Melville is 55 years, worked for 20 years, planned to retire in the next 10 years i.e at the age of 65.

so since he planned to work for the next 10 years;

total duration of work will be; ( 20 + 10) years = 30 years

Average salary per year = $76,908

Total salary in 30 years will be; $76,908 × 30 = $2,307,240

Hospital rate benefits = 2%

So, total annual disability benefit will be;

⇒ 2% of Total salary in 30 years

⇒ 2/100 × $2,307,240

= $46,114.8

Therefore, Herman Melville's annual disability benefit is $46,114.8

4 0

3 years ago

Given that f(x) = x + 3<br> a) Find f(2)

Nataly_w [17]

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

In the above question, it is given that :

$\qquad \sf \boxed{ \sf f(x) = \frac{x + 3}{2} }$

A.) Find f(2) :

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{2 + 3}{2}$

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{5}{2}$

or

$\qquad \sf \dashrightarrow \: f(2) = 0.5$

B.) Find ${ \sf {f}^{-1}(x) }$ :

$\qquad \sf \dashrightarrow \: let \: y = f (x)$

so, we can write it as :

$\qquad \sf \dashrightarrow \: y = \dfrac{x + 3}{2}$

$\qquad \sf \dashrightarrow \: 2y = x + 3$

$\qquad \sf \dashrightarrow \: x = 2y - 3$

Now, put x = ${ \sf {f}^{-1}(x) }$ , and y = x and we will get our required inverse function ~

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(x) = 2x- 3$

C.) Find ${ \sf {f}^{-1}(12) }$ :

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 36- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 33$

7 0

1 year ago