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Nastasia [14]
3 years ago
7

Find the value of c that completes the square 2 + 20r + c

Mathematics
1 answer:
Morgarella [4.7K]3 years ago
7 0

Answer:

c=5

Step-by-step explanation:

5

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Can someone help me with this question?
Karolina [17]

Answer:

D

Step-by-step explanation:

2 x 2 = 4 x PI = 12.56

5 0
3 years ago
Read 2 more answers
Please I need the answer very fast ​
tatyana61 [14]

Answer:

1 - cosΘ

Step-by-step explanation:

using the identity

sin²x + cos²x = 1 , then

sin²x = 1 - cos²x

\frac{sin^20}{1+cos0}

= \frac{1-cos^20}{1+cos0} ← factor 1 - cos²Θ as a difference of squares

= \frac{(1-cos0)(1+cos0)}{1+cos0} ← cancel 1 + cosΘ on numerator/ denominator

= 1 - cosΘ

7 0
2 years ago
Scientists were studying a rare species of parrot. They went to a forest in the year
Brrunno [24]

Answer:

P(t) = 608(1.5)^{t}

Step-by-step explanation:

The number of parrots in t years after 2010 can be modeled by the following function:

P(t) = P(0)(1+r)^{t}

In which P(0) is the number of parrots in 2010 and r is the growth rate, as a decimal.

608 parrots in the forest in 2010.

This means that P(0) = 608

Then

P(t) = 608(1+r)^{t}

When the scientists went back  5 years later, they found 4617 parrots.

This means that P(5) = 4617

We use this to find 1 + r. So

P(t) = 608(1+r)^{t}

4617 = 608(1+r)^{5}

(1+r)^{5} = \frac{4617}{608}

1 + r = \sqrt[5]{\frac{4617}{608}}

1 + r = 1.5

So

P(t) = 608(1.5)^{t}

8 0
3 years ago
Ritz can eat 8 apples in 15 minutes. How long does it take Ritz to eat 16 apples at the same rate?
natka813 [3]
The rate Ritz can eat apples is 8/15 apples per minute.

Set the rate (which is measured apples / minutes) equal to 16 apples / t minutes.

(8/15) = (16/t)

Cross multiply.

16*15 = 8t

t = 20

The answer is 30 minutes. You could solve the problem with basic reasoning as well. 16 is twice the number of apples as 8, so the time it takes to eat them will be double 15 minutes.
5 0
3 years ago
Evaluate the variable expression when a=-4, b=2, c=-3, and d =4. b-3a/bc^2-d​
gtnhenbr [62]

Answer:

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

Step-by-step explanation:

Evaluate:

\dfrac{b-3a}{bc^{2}-d}

When a=-4, b=2, c=-3, and d =4

Solution:

Substitute, a=-4, b=2, c=-3, and d =4 in above expression we get

\dfrac{b-3a}{bc^{2}-d}=\dfrac{2-3(-4)}{2(-3)^{2}-4}\\\\=\dfrac{2+12}{18-4}\\\\

\dfrac{b-3a}{bc^{2}-d}=\dfrac{14}{14}=1

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

6 0
3 years ago
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