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3 years ago

kat,jedric,and roston have 27 total candies kat has 15 jedric has 3 times more than roston how much does jedric have

Mathematics

1 answer:

arsen [322]3 years ago

4 0

I think it is 91 unless i am reading the problem wrong

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Mia's mother gives her $10 per week for allowance Mia gets an additional $3 for every hour she spends watching

nika2105 [10]

Answer:

10+3t=25

Step-by-step explanation:

8 0

3 years ago

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Simplify completely the quantity 4 times x to the third power plus 5 times x to the 2nd power plus 3 times x all over x. (2 poin

ikadub [295]

Answer: 4x² + 5x + 3

<u>Step-by-step explanation:</u>

$\dfrac{4x^3+5x^2+3x}{x}\\\\\\=\dfrac{x(4x^2+5x+3)}{x}\qquad \text{factored out the common term from the numerator}\\\\\\=4x^2+5x+3\qquad \text{canceled out x from numerator and denominator}$

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4 years ago

Tyler buys 3 cans that each contain 3 quarts of blue paint, and he buys 1 can that contains 2 quarts of white paint. What is the

Serga [27]

Tyler has 3 cans of blue paint. Each contains 3 quarts separately. What you need to do is multiply 3 and 3, which would total to 9 quarts. Since you need to include exponents, I´m guessing 3 cubed (3 to the third power) will work. He also buys 1 can of white paint, which only contains 2 quarts. You should add that to the 9 quarts we got from the last steps. I am not sure if this answer is entirely correct, but my guess is 3^3 + 2 = 11 quarts.

6 0

3 years ago

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A base ball is thrown a distance of 60 feet what is its speed if it take 5 seconds to cover the distance

vagabundo [1.1K]

60 ft. / 5 sec = 12 mph.

4 0

3 years ago

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In how many ways can the letters in the word MATHEMATICS be reshuffled so that all consonants appear together?

steposvetlana [31]

Answer:

75600

Step-by-step explanation:

We are given that a word MATHEMATICS

Total letters =11

M repeated 2 times

T repeated 2 times

A repeated 2 times

Total vowels=4

Let MTHMTCS=P

Total number of ways in which MTHMTCS can be arranged= $\frac{7!}{2!2!}$

PAEAI

Total number of ways in which PAEAI can arranged= $\frac{5!}{2!}$

Total number of arrangements when all consonant appear together= $\frac{7!}{2!2!}\times \frac{5!}{2!}$

Total number of arrangements when all consonant appear together= $\frac{7!5!}{2\times 2\times 2}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1\times 5\times 4\times 3\times 2\times 1}{8}=75600$

By using formula ; $n!=n(n-1)(n-2)...2\times 1$

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4 years ago