Help i wiil biran list you

Shia received a large package in the mail. What is a reasonable weight for the large package? A 20 grams B 36 pounds C 52 ML D 6

Harrizon [31]

Answer:

36 pounds

Step-by-step explanation:

Its a "large" package. So it's going to be heavy

4 0

2 years ago

The currency exchange for US dollars (USD) to Canadian dollars (CAD) changes daily. If the rate today is one CAD for every $0.85

Akimi4 [234]

Answer: 9.20 CAD

Step-by-step explanation:

Let this be our conversion factor.

1 CAD = $0.85 USD

We want to go from $7.82 USD to CAD

$7.82USD(\frac{1CAD}{0.85USD}) =9.20CAD$

7 0

3 years ago

At an archery competition, Jamie and Marshall's team tied Shawna and Wyatt's team for the top team score. Team scores are determ

strojnjashka [21]

Given:

Jamie and Marshall's team tied Shawna and Wyatt's team for the top team score.

Marshall scores 12 less than Jamie. Shawna scored twice as much as Marshall. Wyatt scored 4 less than Shawna.

To find:

The individual score of each person.

Solution:

Let x be the score of Jamie.

Marshall scores 12 less than Jamie.

Marshall score = x-12

Shawna scored twice as much as Marshall.

Shawna score = 2(x-12)

Wyatt scored 4 less than Shawna.

Wyatt score = 2(x-12)-4

Now,

Score of Jamie and Marshall's team $=x+(x-12)$

Score of Shawna and Wyatt's team $=2(x-12)+2(x-12)-4$

Jamie and Marshall's team tied Shawna and Wyatt's team

$x+(x-12)=2(x-12)+2(x-12)-4$

$x+x-12=2x-24+2x-24-4$

$2x-12=4x-52$

Isolate x.

$2x-4x=-52+12$

$-2x=-40$

Divide both sides by -2.

$x=\dfrac{-40}{-2}$

$x=20$

So, Jamie score = 20

Marshall score = 20-12

= 8

Shawna score = 2(20-12)

= 2(8)

= 16

Wyatt score = 2(20-12)-4

= 2(8)-4

= 16-4

= 12

Therefore, the scores of Jamie, Marshall, Shawna and Wyatt are 20, 8, 16 and 12 respectively.

7 0

2 years ago

Plot the image of point D under a dilation about the origin (0,0) with a scale factor of 3.

Olenka [21]

Answer:

The correct answer is 0,-6

6 0

2 years ago

once a week you babysit your neighbor’s toddler after school, usually going to a local playground. you notice that each swing on

Natalka [10]

You didn't include the formula.

Given that there is no data about the mass, I will suppose that the formula is that of the simple pendulum (which is only valid if the mass is negligible).

Any way my idea is to teach you how to use the formula and you can apply the procedure to the real formula that the problem incorporates.

Simple pendulum formula:

Period = 2π √(L/g)

Square both sides

Period^2 = (2π)^2 L/g

L = [Period / 2π)^2 * g

Period = 3.1 s
2π ≈ 6.28
g ≈ 10 m/s^2

L = [3.1s/6.28]^2 * 10m/s^2 =2.43 m

I hope this helps you.

6 0

3 years ago