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3 years ago

12

Calculate the monthly payment for a $10,000 auto loan over three years at 5% annual interest.

1 answer:

zhenek [66]3 years ago

4 0

Answer:

Step-by-step explanation:

A

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2. Is the data set "dollars of sales per week” quantitative or qualitative?

Ipatiy [6.2K]

Answer:

this would be quantitative.

Step-by-step explanation:

because its an amount like dollars of sales per week so it would be quantitative or numbers

4 0

3 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista

scZoUnD [109]

The answer to this question
is 60mph

3 0

3 years ago

belka [17]

Answer:

D. 105 students and 150 guests

Step-by-step explanation:

105 students and $1 for each ticket is $105.(105 x 1 = 105)

150 guests and $5 for each ticket is $750.(150 x 5 =750)

$105 + $750 = $855

6 0

2 years ago

Read 2 more answers

Write an equation of the line satisfying the given conditions. Through (2, -5), parallel to 5x 6y + 7

mojhsa [17]

We have to determine the equation of the line passing through the point (2,-5) and parallel to the line $5x = 6y+7$

When two lines are parallel, then the slopes of the two lines are equal.

Equation of line with point $(x_1, y_1)$ and slope 'm' is given by:

$(y-y_1) = m(x-x_1)$

Since, we have to determine the equation of a line with point (2,-5).

So, the equation of the line is : $(y-(-5)) = m(x-2)$

$y+5 = m(x-2)$

Since, the line is parallel to the line $5x = 6y+ 7$

So, $6y +7 = 5x$

$6y = 5x - 7$

$y = \frac{5}{6}x - \frac{7}{6}$

So, slope of the line 'm' is $\frac{5}{6}$ .

Therefore, the equation of the line is:

$(y+5) = \frac{5}{6}(x-2)$

$6y + 30 = 5x - 10$

$6y - 5x = -10-30$

$6y - 5x = -40$

Therefore, $6y - 5x = -40$ is the required equation of the line.

8 0

3 years ago

Consider the initial value problem y′′+36y=2cos(6t),y(0)=0,y′(0)=0. y″+36y=2cos⁡(6t),y(0)=0,y′(0)=0. Take the Laplace transform

Bingel [31]

Recall the Laplace transform of a second-order derivative,

$L(y''(t)) = s^2Y(s)-sy(0)-y'(0)$

and the transform of cosine,

$L(\cos(at))=\dfrac s{a^2+s^2}$

Here, both $y(0)=y'(0)=0$ , so taking the transform of both sides of

$y''(t)+36y(t)=2\cos(6t)$

gives

$s^2Y(s)+36Y(s)=\dfrac{2s}{36+s^2}$

$\implies Y(s)=\dfrac{2s}{(s^2+36)^2}$

4 0

3 years ago