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3 years ago

How are patterns used to solve a problem?

1 answer:

7 0

It can be variety of problems but a x and y plot would be ideal demonstration as you need to make a hypothesis of what you think the pattern for it is then use it for each x and y problem through put the plot

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Lori is three times as old as Michelle. In five years Michelle's age will be one half of Lori's age. How old are Lori and Michel

AURORKA [14]

Answer:

Lori's age is 15 years old and Michelle's age is 5 years old now

Step-by-step explanation:

Let

x-----> Lori's age

y ----> Michelle's age

we know that

$x=3y$ -----> equation A

$(y+5)=\frac{1}{2}(x+5)$ ----> equation B

substitute equation A in equation B and solve for y

$(y+5)=\frac{1}{2}(3y+5)$

$2y+10=(3y+5)$

$3y-2y=10-5$

$y=5$

Find the value of x

$x=3(5)=15$

therefore

Lori's age is 15 years old and Michelle's age is 5 years old now

7 0

3 years ago

Maya had $27. She spent all the money on buying 3 burgers for $x each and 2 sandwiches for $y each. If Maya would have bought 2

defon

Answer:

Option C

The student's conclusion is correct because the solution to the system of equations $3x+2y=27$ and $2x+y=16$ is $(5,6)$

Step-by-step explanation:

Let

x------> the cost of each burger

y-----> the cost of each sandwich

we know that

$3x+2y=27$ -----> equation A

$2x+y=(27-11)$

$2x+y=16$ ----> equation B

Solve the system of equations

Multiply the equation B by $-2$

$-2(2x+y)=-2*16$ -----> $-4x-2y=-32$ ------> equation C

Adds equation A and equation C

$3x+2y=27 \\-4x-2y=-32\\ ----------\\ 3x-4x=27-32\\ -x=-5\\x=5$

Substitute the value of x in the equation B

$2(5)+y=16$

$10+y=16$

$y=6$

The solution of the system of equations is the point $(5,6)$

The price of each burger is $\$5$

The price of each sandwich is $\$6$

6 0

3 years ago

Read 2 more answers

PLEEASSEEEE Answer ASAP. Due in 5 Minutes. 25 points!!!!!!!!!

jonny [76]

Answer:

Here we will use the relationships:

$a^n*a^m = a^{n + m}$

$(a^n)^m = a^{n*m}$

$\frac{a^n}{a^m} = a^{n - m}$

And a number:

$a^n$

is between 0 and 1 if a is positive and larger than 1, and n is negative.

if a is positive and 0 < a < 1, then we need to have n positive such that:

0 < a^n < 1

A) $5^3*5^{-4} = 5^{3 + (-4)} = 5^{-1}$

This is between zero and 1,

B) $\frac{3^5}{3^{-6}} = 3^{5 - (-6)} = 3^{11}$

This is greater than 1, because the exponent is positive.

C) $(\frac{1}{4})^3*( \frac{1}{4})^2 = (\frac{1}{4})^{2 + 3} = (\frac{1}{4})^5$

Because a is smaller than 1, and the exponent is positive, then the expression is between 0 and 1.

D) $\frac{(-7)^5}{(-7)^7} = (-7)^{5 - 7} = -7^{-2}$

The exponent is negative (and pair) then the expression is between 0 and 1.

Remember that when the exponent is pair, we always have that:

(-N)^m = (N)^m

So (-7)^-2 = 7^-2

4 0

3 years ago

In November 2010, an article titled "Frequency of Colds Dramatically Cut with Regular Exercise" appeared in Medical News Today.

disa [49]

Answer:

Part a

For the given study, the explanatory variable or independent variable is given as regularity or frequency of exercise. This variable is classify as categorical variable because variable is divided into two categories such as whether participant exercise 5 or more days a week or not.

Part b

For the given study, the response variable or dependent variable is given as frequency of colds. This variable is classified as quantitative variable because we measure the quantities or frequency of number of colds.

Part c

A confounding variable for this research study is given as incidence of upper respiratory tract infections that provides an alternative explanation for the lower frequency of colds among those who exercised 5 or more days per week, compared to those who were largely sedentary. This confounding variable is categorical in nature.

5 0

3 years ago

Anybody good at solving solutions?? Need done ASAP

Leto [7]

I am pretty sure that B is the answer

4 0

2 years ago