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zepelin [54]
2 years ago
6

Students in Ms. Singh's class had the following scores on their last quiz.

Mathematics
1 answer:
Svetllana [295]2 years ago
8 0
the median of her 20 students is 82.5
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What is the length of DE?
g100num [7]

Answer:

Step-by-step explanation:

Yes

Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Given: ∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Prove: ∠A ~= ∠C.

 Statements Reasons

1. ∠A and ∠B are complementary, and ∠C and ∠B are complementary. Given

2. m∠A + m∠B = 90º , m∠C + m∠B = 90º Definition of complementary

3. m∠A = 90 º - m∠B, m∠C = 90º - m∠B Subtraction property of equality

4. m∠A = m∠C Substitution (step 3)

5. ∠A ~= ∠C Definition of ~=

6 0
3 years ago
What is the value of p+9 when p=16​
Tasya [4]

Answer:

25 :)

Step-by-step explanation:

well, if it's p+9 and you know that p=16 then you plug 16 in for p so the equation will be 16+9 and then you solve from there which is 25

4 0
2 years ago
(10x-4)+(-7+x)<br> Find the sum<br><br> Please Help!!!!
12345 [234]
The answer is 11x-11
5 0
2 years ago
Which of the following options results in a graph that shows exponential growth?
Galina-37 [17]
I know for a fact B is wrong, so you can mark that off. I got C.
6 0
2 years ago
Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
Marizza181 [45]

If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams

The half life in years = 24100

Consider the quantity of the radio active isotope remaining

y = Ce^{kt}

When t = 1000 the y = 1.2

y = C/2 when t = 1599

Substitute the values in the equation

C/2 = Ce^{24100k}

Cancel the C in both side

1/2 = e^{24100k}

Here we have to apply ln to eliminate the e terms

ln (1/2) = 24100k

k = ln(1/2) / 24100

k = -2.87× 10^-5

To find the initial value we have to substitute the value of k and y in the equation

1.2 = Ce^{1000 ×  -2.87× 10^-5}

C = 1.2 / e^(-0.0287)

C = 2.16 gram

Hence, the initial quantity of the radioactive isotope is 2.16 gram

Learn more about half life here

brainly.com/question/4318844

#SPJ4

7 0
1 year ago
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