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Dafna11 [192]
3 years ago
15

When solving for a variable in an equation, it is important to complete a 'goals' list so you can..."

Mathematics
1 answer:
DanielleElmas [232]3 years ago
3 0
You must follow Mathematical properties and apply it with your logic, so E is the answer
You might be interested in
Which shows one way to determine the factors of x3 – 9x2 + 5x – 45 by grouping?
REY [17]
X³ - 9x² + 5x - 45

x²(x - 9) - 5(x - 9) ⇒ x³ - 9x² - 5x + 45 INCORRECT
x²(x + 9) - 5(x + 9) ⇒ x³ + 9x² - 5x - 45 INCORRECT

x(x² + 5) - 9(x² + 5) ⇒ x³ + 5x - 9x² - 45 CORRECT
It just needs to be rearranged.

x(x² - 5) - 9(x² - 5) ⇒ x³ - 5x - 9x² + 45 INCORRECT
4 0
4 years ago
Aaron kicked a soccer ball with an initial velocity of 39 feet per
marysya [2.9K]

Answer:

29.6 feet

Step-by-step explanation:

6 0
3 years ago
Find the length of the radius. Assume that lines which appear to be tangent are tangent.
aleksley [76]

The Answer Is 7.5

Tangent and 8 makes a right triangle, which is great because you could find the diameter and divide it by 2 to find the radius.

64+d^2=289

d^2=225

d=15

15/2=7.5

3 0
3 years ago
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

4 0
4 years ago
3.14*16.8*16.8*16.9
vitfil [10]

Answer: 3.14 *16.8*16.8*16.9=14977.3478

14977.3478= 14,977.35

6 0
3 years ago
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