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pav-90 [236]
3 years ago
8

Okay here is another math problem I'll give a brainliest for the right answer!

Mathematics
1 answer:
Sliva [168]3 years ago
4 0
To find the total area of this figure, you have to find the area of each of the separate figures and then add those 2 answers together. Your work should look like this:

2 x 2 = 4

8 x 8 = 64

64 + 4 = 68

Your answer should be 68 ft.^2
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Please answer this I’ll make you brainlienst!!!
igor_vitrenko [27]

Answer:

? = 2

Step-by-step explanation:

<u>Step 1:  Substitue 1 for A and 3 for B</u>

<u />\frac{2(1)(3) + 2}{2(1)(3) - 2} = ?

\frac{6 + 2}{6 - 2} = ?

\frac{8}{4} = ?

\frac{2}{1} = ?

2 = ?

Answer: ? = 2

5 0
3 years ago
Read 2 more answers
Which statement is true about the residual plot?
svetlana [45]

Answer: the first on

Step-by-step explanation:

3 0
3 years ago
Mary borrowed $11,000 at 5% interest for 4 years. What was the total interest? A. $22 B. $220 C. $2,200 D. $22,000
pantera1 [17]
<span>$11,000 x 4 x 0.05 = 2200

answer is </span><span>C. $2,200</span>
6 0
3 years ago
Read 2 more answers
Are these two answers correct? Please help, thanks!
Reika [66]
Let's help Julia to solve this problem. Given the term:

\frac{sec^{2}( \theta)-1}{cot^{2}( \theta)+1}

We need to simplify it. So, we will take a look at the identities above, thus, from the second identity:

sec^{2}( \theta)-tan^{2}(\theta) = 1

∴ sec^{2}( \theta)-1=tan^{2}(\theta)

And from the five equation:

1 = csc^{2}( \theta)-cot^{2}(\theta)

∴ cot^{2}(\theta)+1= csc^{2}( \theta)

Substituting these identities in the term:

\frac{tan^{2}( \theta)}{csc^{2}( \theta)} =  sin^{2}( \theta)tan^{2}( \theta)

In fact, your answer is correct.
7 0
3 years ago
Read 2 more answers
In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that
leonid [27]

Given:

Number of students who has a cat and a dog = 5

Number of students who has a cat but do not have a dog = 11

Number of students who has a dog but do not have a cat = 3

Number of students who neither have a cat nor a dog = 2

To find:

The probability that a student has a cat given that they do not have a dog.

Solution:

Let the following events:

A = Student has a cat

B = Do not have a dog

Total number of outcomes is:

5+3+11+2=21

The probability that a student has a cat but do not have a dog is:

P(A\cap B)=\dfrac{11}{21}

The probability that a student do not have a dog is:

P(B)=\dfrac{11+2}{21}

P(B)=\dfrac{13}{21}

The conditional probability is:

P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}

P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}

P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}

Therefore, the probability that a student has a cat given that they do not have a dog is \dfrac{11}{13}.

5 0
3 years ago
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