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OLEGan [10]
3 years ago
12

HELP ASAP!!! Find the value of 'A' in the set of complementary angles.

Mathematics
1 answer:
dem82 [27]3 years ago
7 0

answer:

180 - 63 = 117

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Dyami lives x miles from the school. Karla lives 2/3 further than that from the school.
Hatshy [7]

Answer:

Where is the schools coordinates subtract two thirds from whatever the schools coordinates are and you will have your answer for Dyami

Step-by-step explanation:

3 0
2 years ago
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Which shows all the like terms in the expression?
GenaCL600 [577]
4x-3+7x+1

Ones with a variable: 4 & 7
Only whole number: -3 & 1

The answer is option two: -3 and 1; 4x and 7x
7 0
3 years ago
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HELP PLEASE 80 POINTS
alexgriva [62]

Answer:

  • x = 15/p
  • x = -3

Step-by-step explanation:

<h3>Given</h3>
  • 4(px + 1)=64
<h3>Find</h3>
  • Solve for x
  • Find x when p = -5
<h3>Solution</h3>
  • 4(px + 1) = 64
  • 4(px + 1)/4 = 64/4
  • px + 1 = 16
  • px = 15
  • x = 15/p

<u>When p = -5, substitute p:</u>

  • x = 15/p
  • x = 15/(-5) = -3
4 0
3 years ago
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PLEASEEEE HELP I WILL GIVE BRAINLIEST!!
I am Lyosha [343]

Answer:

The average is 5.41666667 degrees.

Step-by-step explanation:

3 0
3 years ago
A population of bacteria begins with 1500 bacteria and grows to 4500 in one hour.
dezoksy [38]

The function that represents the growth of this culture of bacteria as a function of time is; P = 1500e^(1.0986t)

<h3>How to calculate Exponential Growth?</h3>

The formula for exponential growth is;

P = P₀e^(rt)

where;

P = current population at time t

P₀ = starting population

r = rate of exponential growth/decay

t = time after start

Thus, from our question we have;

4500 = 1500 * e^(r * 1)

4500/1500 = e^r

e^r = 3

In 3 = r

r = 1.0986

Thus, the function that represents the growth of this culture of bacteria as a function of time is;

P = 1500e^(1.0986t)

For the culture to double, then;

P/P₀ = 2. Thus;

e^(1.0986t) = 2

In 2 = 1.0986t

t = 0.6931/1.0986

t = 0.631 hours

Read more about Exponential Growth at; brainly.com/question/27161222

#SPJ1

4 0
2 years ago
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