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3 years ago

3/5 + -2/7 in simplest form

Mathematics

1 answer:

Lemur [1.5K]3 years ago

5 0

In the attachment this is how i did it! ~Best of Luck~

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Find the domain of y=3x+3 square root

ad-work [718]

Answer:

Answer is given below

Step-by-step explanation:

$y = \sqrt{3x + 3} \\$

when x=2,11,........

domain is 2 and 11

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

8 0

3 years ago

The area (A of a circle with a radius of ris given by the formula A = 772 and its diameter (d) is given by d = 27. Arrange the e

lions [1.4K]

Answer:

A = πr² ← r is the radius and r = \frac{1}{2} d ← d is the diameter

Hence

π (\frac{1}{2} d)² = A, that is

\frac{d^2\pi }{4} = A ( multiply both sides by 4 )

d²π = 4A ( divide both sides by π )

d² = \frac{4A}{\pi } ( take the square root of both sides )

d = \sqrt{\frac{4A}{\pi } }

4 0

3 years ago

A carpenter worked 34 hours a week for half a year. If her hourly wage was $20, how much did she earn during this time period?

Mrrafil [7]

Keep in mind units and it will make sense
20(34)(4*6) = ?

4 0

3 years ago

The amount y (in grams) of the radioactive isotope fermium-253 remaining after t hours is y=a(0.5)t/72, where a is the initial a

almond37 [142]

Answer:

Per hour decay of the isotope is 0.96%.

Step-by-step explanation:

Amount of radioactive element remaining after t hours is represented by

$y=a(0.5)^{\frac{t}{72}}$

where a = initial amount

t = duration of decay (in hours)

Amount remaining after 1 hour will be,

$y=a(0.5)^{\frac{1}{72} }$

y = 0.9904a

So amount of decay in one hour = a - 0.9904a

= 0.0096a gms

Percentage decay every hour = $\frac{\text{Amount of decay}}{\text{Initial amount}}\times 100$

= $\frac{0.0096a}{a}\times 100$

= 0.958 %

≈ 0.96 %

Therefore, per hour decay of the radioactive isotope is 0.96%.

7 0

3 years ago

HELPPP THIS IS DUE SOON! I WILL MARK BRAINLIEST

expeople1 [14]

Answer:

7.3

Step-by-step explanation:

For this question, you must use the distance formula. The distance formula is based around Pythagorean Theorem, so you will see some similarities.

Distance = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$

X2 is 7 (X coordinate of school)

X1 is 5 (X coordinate of friend's house)

7 - 5 = 2

2^2 = 4

Y2 is 7 (Y coordinate of school)

Y1 is 0 (Y coordinate of friend's house)

7 - 0 = 7

7^2 = 49

4 + 49 = 53

$\sqrt{53}$ = 7.2801...

Round to the nearest tenth...

7.3

4 0

2 years ago

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