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Vinil7 [7]
4 years ago
12

Write the product in standard form. ( 4 + 7i)( 7 + 7i)

Mathematics
1 answer:
Helen [10]4 years ago
7 0
The answer is a) -21 - 77i
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Prove <5 is supplementary to <6 using theorems.
babunello [35]

Answer:

Theorem 10.5: If two parallel lines are cut by a transversal, then the exterior angles on the same side of the transversal are supplementary angles.

6 0
3 years ago
Please help me A.S.A.P.
Vikentia [17]

Answer:

700 Dollars

Step-by-step explanation:

The expression for solving this is I = prt

So lets start with P, that is our principal or amount of money we start with, and we want to multiply it by our rate, or r , which is 3.5, so we would do 5000 x 0.035 to get 175, then we multiply that by our time or t, which is 4 years, so 175 x 4 = 700 dollars :D

5 0
3 years ago
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a video game arcade offers a yearly membership with reduced rates for game play. A single membership costs $60 per year. Game to
7nadin3 [17]

slope is $0.10            <em>($1.00 per 10 tokens = $0.10)</em>

y-intercept is $60    <em>($60 is the annual fee)</em>

y = .10x + 60            <em>(y = mx + b)</em>  

domain is x ≥ 0        <em>(you can't buy a negative number of token)</em>

range is y ≥ 60        <em>(input the domain (x ≥ 0) to find the range)</em>

Answer: the y-intercept of the function is $60

6 0
3 years ago
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Which list shows these numbers in order from least to greatest?
ira [324]

Answer:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15...

3 0
4 years ago
Given that f(a+b)= f(a) + f(b) and f(x) is always positive, what os the value of f(0)?
FrozenT [24]
This is a strange question, and f(x) may not even exist. Why do I say that? Well..

[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).

[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!

Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.

To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
4 0
3 years ago
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