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FinnZ [79.3K]
3 years ago
5

Guess the car hint**Uncommon**

Mathematics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

Albar sonic?

Step-by-step explanation:

My grandpa showed me a picture of a car that looked like this. He said it was a Albar sonic.

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Deshi has 427 marbles and Elaheh has 394 marbles Deshi wants to give 125 marbles to his little brother how many total marbles wi
hjlf

Answer: Deshi will have 302 marbles  and Elaheh will have 519.

Step-by-step explanation:

Deshi's Marbles will subtract 125.

427 - 125 = 302.

Elaheh's Marbles will add 125.

394 + 125= 519.

3 0
2 years ago
Which inequality can be used to explain why these three segments cannot be used to construct a triangle?
ryzh [129]

The answer choice which explains that the three segments cannot be used to construct a triangle is; AC + CB < AB.

<h3>Which inequality explains why the three segments cannot be used to construct a triangle?</h3>

Since, It follows from the triangle inequalities theorem that sum of the side lengths of any two sides of a triangle is greater than the length of the third side.

Hence, since the sum of sides AC + CB is less than AB, it follows that the required inequality is; AC + CB < AB.

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7 0
1 year ago
2x + y = -2
cupoosta [38]
Answer is A
if x+y=5
y= -x+5
2x + (-x+5) =-2
2x -x +5 = -2
x+5 = -2
x=-7

5 0
3 years ago
Read 2 more answers
A Farmer plants the same amount everyday, adding up to 1 3/4 acres at the end of the year. If the year is 1/2 over how many acre
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Answer:

1/45 acres the farmer planted

8 0
3 years ago
Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
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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

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7 0
1 year ago
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