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1 year ago

The solution to the given system of linear equations lies in which quadrant?

Mathematics

2 answers:

goblinko [34]1 year ago

8 0

Answer:

The solution lies in quadrant IV.

Step-by-step explanation:

Rina8888 [55]1 year ago

5 0

X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation.
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant

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Jeannie needs 2 cups of milk to make her homemade brownies. How many fluid ounces does Jeannie need?

vovikov84 [41]

Answer:

16 fluid ounces=2 cups

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Step-by-step explanation:

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1 year ago

Read 2 more answers

Devin borrowed $1,058 at 13 percent for nine months. What will he pay in interest?

Viktor [21]

Devin borrowed $1,058 at 13 percent for nine months.

We have to calculate the interest paid.

Interest = $\frac{P \times R \times T}{100}$

Substituting the values of

Principal = $1058

Rate = 13%

Time = 9 months = $\frac{9}{12}$ year

Interest = $\frac{1058 \times 13 \times 9}{12 \times 100}$

Interest = 103.155

= 103.16

So, Devin will pay 103.16 as the interest.

Therefore, Option A is the correct answer.

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2 years ago

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Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y

zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

8 0

2 years ago

A new obstacle course is home to several new types of challenging obstacles. One such obstacle is traversing a 15-stair staircas

Tresset [83]

Answer: $22\frac{1}{2}\text{ Feet}$

Step-by-step explanation:

Here, the vertical space between two consecutive stairs in first 10 stairs = $1\frac{1}{3}\text{ feet}$

⇒ The total vertical distance for 10 stairs = $10\times 1\frac{1}{3}=10\times \frac{4}{3}=\frac{40}{3}\text{ feet}$

Also, The vertical space between two consecutive stairs for last 5 stairs = $1\frac{3}{4}\text{ feet}$

⇒ The total vertical distance for 10 stairs = $10\times 1\frac{1}{3}=5\times \frac{7}{4}=\frac{35}{4}\text{ feet}$

Thus, The total vertical distance = Total vertical space for the first 10 stairs + Total vertical space for the last 5 stairs

= $\frac{40}{3} + \frac{35}{4}$

= $\frac{160+105}{12}$

= $\frac{265}{12}$

= $22\frac{1}{2}\text{ Feet}$

4 0

1 year ago

For a recent season in college football, the total number of rushing yards for that season is recorded for each running back. Th

Galina-37 [17]

Answer:

The standard deviation of the number of rushing yards for the running backs that season is 350.

Step-by-step explanation:

Consider the provided information.

The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards.

Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.

Use the formula: $z=\frac{x-\mu}{\sigma}$

Here z is 2.42 and μ is 790, substitute the respective values as shown.

$2.42=\frac{1637-790}{\sigma}$

$\sigma=\frac{847}{2.42}$

$\sigma=350$

Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.

4 0

2 years ago