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Gnom [1K]
2 years ago
6

The sum of 5 consecutive even numbers is 310.

Mathematics
1 answer:
RideAnS [48]2 years ago
8 0

Answer:

your answer would be 186

Step-by-step explanation:

310 / 5 = 62

62 x 3 = 186

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X+10=1/3(5x+10)<br><br> How do you solve this equation?
Nikitich [7]

Answer:

x = 10

Step-by-step explanation:

First, subtract 10 from both sides

Next, Simplify

Then, subtract 5/3x from both sides

After, simplify


Lastly, multiply both sides by 3


Then finally simplify for your final answer of x = 10


Hope this helps! :)

3 0
2 years ago
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Find f(-2) if f(x) = x + x + 7. f(-2) =
Lynna [10]

Answer:

f(-2) = 9

Step-by-step explanation:

Given that,

f(x) = x² + x + 7

We need to find the value of f(-2).

Put x = -2 in the given function.

f(-2) = (-2)² + (-2) + 7

= 4-2+7

= 9

Hence, the value of f(-2) is equal to 9.

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3 years ago
(3/4)3 write as improper fraction and mixed number
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15/4, 3 3/4, and 3.75
5 0
3 years ago
"A class of 30 students took midterm science exams. 20 students passed the chemistry exam, 14 students passed physics, and 6 stu
katen-ka-za [31]
The answer is the letter B 
4 0
3 years ago
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We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0
2 years ago
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