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netineya [11]
2 years ago
12

Find the solution to the system of equations using graphing y = -7/4x - 4 y = - 1/4x + 2

Mathematics
1 answer:
dlinn [17]2 years ago
4 0

Answer: look at the picture

Step-by-step explanation: Hope this help :D

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the lateral surface area of right circular cylinder is 120 pi cm squared and the circumference is 12 cm find the height ​
konstantin123 [22]

Answer:

h=31.4 cm

Step-by-step explanation:

A=lateral surface area

c=circumference

h=height

A= c*h

solving for h we get:

h=A/c=120pi/12=10pi

6 0
2 years ago
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Solve the equation 3x = 9.5 Show and justify each step you take to solve the equation . 3x = 9 Write the equation 9 Divide each
zimovet [89]

Answer:

3x=9.5, you would divide 3 on each side to get x by itself, so (3x)/3=9.5/3, which would lead to be x=9.5/3 or x= 3 and 1/6th, 3 + 1/6

Step-by-step explanation:

4 0
2 years ago
How much will be left after 24 hrs if it’s half life is 6 hours?
Aleks [24]
Hello there! A half-life is basically a decay of 50%. We will find the amount left after one day by exponential decay. The formula for exponential decay is A(1 - r)^t, where A = initial amount, r = rate (in decimal form), and t = time (or in this case, amount of half lives). The half-life is 6 hours and 24 goes into 6 four times. That's 4 half lives. Let's start by subtract 50% (0.5) from 1. 1 - 0.5 is 0.5. Now, we will use 0.5 and raise it to the 4th power, because the 4th power represents 4 half lives. 0.5^4 is 0.0625. Leave this number on the calculator. Do not round. Multiply that number by 448 to find the amount left. When you do, you get 28. There. 28 grams will be left after 24 hours.
3 0
3 years ago
What is the length of one side of a cube that has a surface area of 1,734 square mm?
jarptica [38.1K]

Answer:

a = 17

Step-by-step explanation:

1,734 = 6a^2

1,734/6 = 6a^2/6

\sqrt{289} = \sqrt{a^2}

a = 17

7 0
2 years ago
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Solve the equation for all real solutions.<br> 8w2 + 2w + 2 = 3
fredd [130]

Answer:

Step-by-step explanation:

w=  1 /4   or w=  −1/ 2

3 0
3 years ago
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