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3 years ago

Draw the grid of 3y- 4x=12

Mathematics

1 answer:

mylen [45]3 years ago

8 0

This is the grid of 3y - 4x = 12

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Use compatible numbers to estimate the quotient.<br> 486/6

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The answer is 81. You just have to do the division in the traditional way.

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4 years ago

Natasha wants to find out if the neighborhood supports lowering the speed limit on the street in front of her school. Which is a

Daniel [21]

The answer is
Thirty residents who live within a 2 mile radius of natashas school.

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2 years ago

Write an algebraic expression for the word expression: <br> 15 divided by the sum of "d" and 3

Yuki888 [10]

the answer is

15/(3+d)

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4 years ago

Ashley, Bob, Claire, and Daniel are among 13 students who entered a lottery to win a free vacation to Paris.

AlladinOne [14]

Answer:

0.0014 = 0.14% probability that Ashley, Bob, Claire, and Daniel will be chosen.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the students are chosen is not important, so the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes:

4 students from a set of 4(Ashley, Bob, Claire, and Daniel). So

$D = C_{4,4} = \frac{4!}{4!(4-4)!} = 1$

Total outcomes:

4 students from a set of 13(number of students in the lottery). So

$T = C_{13,4} = \frac{13!}{4!(13-4)!} = 715$

Probability:

$p = \frac{D}{T} = \frac{1}{715} = 0.0014$

0.0014 = 0.14% probability that Ashley, Bob, Claire, and Daniel will be chosen.

7 0

3 years ago

Prove that an = 4^n + 2(-1)^nis the solution to

olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute $a_n=4^n+2(-1)^n$ in the equation $a_n=3a_{n-1}+4a_{n-2}$ the equality is true.

Let's substitute first in the right hand side:

$3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})$

Now we use the distributive laws. Also, note that $(-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n}$ (this also works when the power is n-2).

$=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}$

$=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}$

$=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n$

then the sequence solves the recurrence relation.

4 0

3 years ago