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yan [13]
2 years ago
7

What method would you choose to solve the equation 2x2 – 7 = 9? Explain why you chose this method.

Mathematics
2 answers:
arlik [135]2 years ago
8 0

Sample Response/Explanation: I would use the square root property of equality. Because there is no x term, I would just add and divide to get x2 by itself. Then I would take the square root.

oksano4ka [1.4K]2 years ago
7 0

I would add 7 to get ...

... 2x² = 16

then divide by 2 to get

... x² = 8

Now, I would take the square root, recognizing that both positive and negative values will solve the problem.

... x = ±√8

The root can be simplified, so we have ...

... x = ±2√2

_____

The above method seemed the most obvious method to me. One could use the quadratic formula, but that seems to involve more operations.

... 2x² -16 = 0 . . . . subtract 9

... x = (-0 ± √(0² -4·2·(-16)))/(2·2) . . . . put the coefficients into the formula

... x = ±(√128)/4 = ±√8 = ±2√2 . . . . . simplify the result

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Find the volume of a right circular cone that has a height of 17.7 m and a base with a diameter of 18.2 m. Round your answer to
Korvikt [17]

Steps:

1: Use the formula V=πr2h 3 to calculate the volume for Right circular cone

2: H is the height,  R is the Radius

3: Use 17.7 to calculate the Height.

4: Use  18.2  to calculate the Radius

Description:

Radius: 18.2

Height: 17.7

Shape: Right circular cone

Solved for volume

Formula: V=πr2h 3

Answer: V≈6139.66

Please mark brainliest

<em><u>Hope this helps.</u></em>

4 0
2 years ago
Which of the following functions is graphed below?
LuckyWell [14K]
The answer to this is D
5 0
1 year ago
What is the sum of square root of -2 and square root of -18
Tasya [4]
Answer is B
4 square root of 2i

7 0
2 years ago
You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in
emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

j(t)=95t-9t^2

j'(t)=95-18t

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

t=\frac{95}{18}

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

j(5.28)=95(5.28)-9(5.28)^{2}

j(5.28)=250.69ft

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

j(t)=95t-9t^{2}

j'(t)=95-18t

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3)  yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

5 0
2 years ago
"A 10-gram marble has a speed that is 5 times as fast as that of a 100-gram marble. Both marbles leave the table at the same tim
Ostrovityanka [42]
If the 10-gram marble is going faster it is possible that the momentum of the marble will make if go farther than the 100-gram marble
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