Ask question

Ask question

All categories

3 years ago

7

You exercised 24 hours each month for a year. How many hours did you exercise by the end of the year? You may be able to do the

math mentally thanks to expanded notation and the Distributive Property.

2 answers:

dedylja [7]3 years ago

7 0

Answer:

288 hours.

Step-by-step explanation:

You exercise 24 hours in 1 month.

There are 12 months in one year.

24 * 12 = 288

Therefore, you exercised 288 hours in one year.

alexgriva [62]3 years ago

4 0

Answer:

Calculation with Distribution

24 (1+1+1+1+1+1+1+1+1+1+1+1)

24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1

24+24+24+24+24+24+24+24+24+24+24+24

288 hours

You did 288 hours of exercise.

Calculation without Distribution

24(1+1+1+1+1+1+1+1+1+1+1+1)

24(12)

288 hours

You did 288 hours of exercise.

Step-by-step explanation:

You might be interested in

Some help me asp i need help to this question

Alexus [3.1K]

Answer:

X=11 Y=9

Step-by-step explanation:

6 0

3 years ago

Last year at this time we had 250 employees. Due to attrition , we lost 12% of our workforce. How many employees are working now

Brums [2.3K]

You have 220 workers now

3 0

3 years ago

Read 2 more answers

PLease answer !!! Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all

Xelga [282]

Answer:

1. $(A,B) = (3,-2)$

2. The values of t are: -3, -1

Step-by-step explanation:

Given

$\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

$|t| = 2t + 3$

Required

Solve for the unknown

Solving $\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

Take LCM

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}$

Expand the denominator

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}$

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}$

Both denominators are equal; So, they can cancel out

$x + 7 = A(x+1) + B(x-2)$

Expand the expression on the right hand side

$x + 7 = Ax + A + Bx - 2B$

Collect and Group Like Terms

$x + 7 = (Ax + Bx) + (A - 2B)$

$x + 7 = (A + B)x + (A - 2B)$

By Direct comparison of the left hand side with the right hand side

$(A + B)x = x$

$A - 2B = 7$

Divide both sides by x in $(A + B)x = x$

$A + B = 1$

Make A the subject of formula

$A = 1 - B$

Substitute 1 - B for A in $A - 2B = 7$

$1 - B - 2B = 7$

$1 - 3B = 7$

Subtract 1 from both sides

$1 - 1 - 3B = 7 - 1$

$-3B = 6$

Divide both sides by -3

$B = -2$

Substitute -2 for B in $A = 1 - B$

$A = 1 - (-2)$

$A = 1 + 2$

$A = 3$

Hence;

$(A,B) = (3,-2)$

Solving $|t| = 2t + 3$

Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;

$t = 2t + 3$ and $-t = 2t + 3$

Solving $t = 2t + 3$

Make t the subject of formula

$t - 2t = 3$

$-t = 3$

Multiply both sides by -1

$t = -3$

Solving $-t = 2t + 3$

Make t the subject of formula

$-t - 2t = 3$

$-3t = 3$

Divide both sides by -3

$t = -1$

<em>Hence, the values of t are: -3, -1</em>

7 0

2 years ago

Read 2 more answers

A. P(5 < X < 9)<br>B. P(X >7)<br>

enot [183]

i think it is A. P(5<X<9)

8 0

3 years ago

please help i rly need it my math teacher gave me over 30 assignments and a project that are all due tomorrow today and i want t

Sedaia [141]

Answer:

x= -14

Step-by-step explanation:

-14 subtracted by 2 is -16, since its absolute it means its positive, meaning its 16 greater than 14, I may be wrong so just to let you know. Depends on how the teacher wants it to be.

8 0

3 years ago