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horsena [70]
3 years ago
10

Write -2i + (9 - 3i) - (6 - 10i) as a complex number in standard form. 3 - 5i 3 + 5i 3 + 16i 12 - 16i

Mathematics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

3 + 5i

Step-by-step explanation:

Given

- 2i + (9 - 3i) - (6 - 10i) ← distribute parenthesis

= - 2i + 9 - 3i - 6 + 10i ← collect like terms

= 3 + 5i

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Apply the distribution
kicyunya [14]

Answer:

20(3m-2)

Step-by-step explanation:

In order to factor, you have to look for the greatest common factor, or what the biggest number could the two numbers be divided by.

In this case, both numbers can be divided by 20, and so we put the 20 outside of the parentheses.

This leaves with 3m and -2, which are left inside the parentheses.


Hope this helps :)


8 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
Each guest at casandras party had 3 turns to act out worlds during a game. They began with a list of 50 words. When they finishe
Veronika [31]

Answer:

16 fits but not evenly

Step-by-step explanation:

16 times 3 equals 48 and 17 times 3 equal is 51 so someone didnt go 3 times.

7 0
3 years ago
He has 9 cans of white paint and 20 cans of blue paint.
Liula [17]

Answer:

The answer to the problem is 1/2

3 0
3 years ago
Can some one please help me to teach this to my daughter
EastWind [94]
28mi/1 hr = 28 mi/60 min = 1 mi/(60/28) min = 

28 mi/hr = 28 mi/60 min since there are 60 min in 1 hr
1 mi/(60/28) min since you divide top and bottom of 28/60 by 28 to get
1 mi/(60/28) min = 1mi/(15/7) min = 1 mi/ 2 1/7 min
3 0
3 years ago
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