All categories

3 years ago

Carrie started work at 8 a.m. and left at 1:30 p.m. How many hours of work should she record on her time card for the day?

2 answers:

5 0

Start: 8 a. m.
Finish: 1 : 30 p.m. = 13 hours 30 minutes
13 h 30 min - 8 h = 5 h 30 min
She should record 5 hours and 30 minutes on her time card for the day.

Aloiza [94]3 years ago

4 0

5 hours 30 minutes if im right

You might be interested in

Why can you NOT combine 3x and 5x^2 ?

diamong [38]

Although they are the same variable, you cannot add two variables raised to different powers. 3x^1 + 5x^2 cannot work, but 3x^2 + 5x^2 can. Also, if you are multiplying them, they would combine to be 15x^3, as 3 and 5 (The coefficients) multiply together and x^2 times x^1 = x^3.

I hope that helps.

7 0

3 years ago

write the equation of the line that is parallel to y = 4x -7 and passes through the point (-2, -5). show your work.

natita [175]

Answer:

y = 4x + 3

Step-by-step explanation:

To do this, put the new line into the form y = mx + b where m is the slope and b is the y-intercept.

When two lines are parallel, they have the same slope. Since, the slope of the first line is 4, that means the slope of the second line is also 4.

So far, the equation is y = 4x + b.

Then, since you know (-2, -5) is a solution, x = -2, y = -5 is a solution so you can plug this into the equation.

This would be:

-5 = 4 * -2 + b

-5 = -8 + b

3 = b

This means the equation is y = 4x + 3.

7 0

2 years ago

PLEASE HELP ME I'M GIVING 20PTS AND MARKING BRAINLIEST!!!!

pishuonlain [190]

Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

$\cos^2{\theta} = 1 - \frac{1}{9}$

$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

#SPJ1

5 0

1 year ago

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago

What is the missing exponent in the equation 2? = 16?

IRISSAK [1]

Answer:

Step-by-step explanation:

2x2x2x2 = 16

5 0

3 years ago

Read 2 more answers