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4 years ago

9

A corporate team building event costs $2 for every attendee. If there are 8 attendees how much will the corporate team building

event Cost?

1 answer:

lesantik [10]4 years ago

7 0

Answer:

16$

Step-by-step explanation:

8 times 2 is 16

I hope this helps!!!

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Find the measure of the indicated arc

dimaraw [331]

Answer:

arc of WXY= 224°

Step-by-step explanation:

measure of angle C is half of the WXY arc

this means WXY×1/2=C

112°=1/2 WXY (cross production)

WXY=2×112°

WXY=224 °

5 0

3 years ago

Zeke is shipping clerk for a large business. Today he spent 90 minutes preparing boxes for shipping. One box weighed 10 pounds a

Ugo [173]

10 + (7 * 3.5) = 34.5

The total weight of the boxes is 34.5 lb.

Thanks for your question! Don't forget to rate 5 stars and give me the brainliest answer! Then, I can help you with all your problems! ^-^ ~

4 0

4 years ago

An investment of $8,000 earns interest at an annual rate of 7% compounded continuously. Complete parts (A) and (B) below. Click

Sergeeva-Olga [200]

Answer:

A. $\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

B. $\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

Step-by-step explanation:

Given that:

An investment of Amount = $8000

earns at an annual rate of interest = 7% = 0.07 compounded continuously

The objective is to :

A) Find the instantaneous rate of change of the amount in the account after 2 year(s).

we all know that:

$A = Pe^{rt}$

where;

$A = (8000) \ e ^{0.7t}$

The instantaneous rate of change = $\dfrac{dA}{dt}$

$\dfrac{dA}{dt} = \dfrac{d}{dt}(8000 \ e ^{0.07t} )$

$= 8000 \dfrac{d}{dt}e^{0.07 \ t}$

$\dfrac{dA}{dt}= 8000 (0.07)e^{0.07 \ t}$

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 2 years; the instantaneous rate of change is:

$\dfrac{dA}{dt}|_{t=2}= 560 e^{0.07 \times 2}$

$\mathtt{\dfrac{dA}{dt}|_{t=2}=644.15}$

(B) Find the instantaneous rate of change of the amount in the account at the time the amount is equal to $12,000.

Here the amount = 12000

$12000 = (8000)e^{0.07 \ t}$

$\dfrac{12000 }{8000}= e^{0.07 \ t}$

$1.5= e^{0.07 \ t}$

㏑(1.5) = 0.07 t

0.405465 = 0.07 t

t = 0.405465 /0.07

t = 5.79

$\dfrac{dA}{dt}= 560 e^{0.07 \ t}$

At t = 5.79

$\dfrac{dA}{dt}|_{t = 5.79}= 560 e^{0.07 \times 5.79}$

$\mathtt{\dfrac{dA}{dt}|_{t = 5.79}= 839.86 }$

7 0

3 years ago

The driving distance between Manchester and London is 195 miles. Farris intends to travel from Manchester to London by coach. Th

sammy [17]

Answer:

Faris' arrival time = 3:30 pm + 3 h 54 mnt = 7:24 pm

Step-by-step explanation:

Driving distance between Manchester to London = 195 miles

Coach left Manchester = 3.30 pm

Average speed of coach = 50 mph

Total time taken by the coach to reach the Manchester to London = 195÷50 = 3.9 h

0.9 h = 0.9×60 mnt = 54 mnt

Now,

3.9 h = 3 h 54 mnt

Hence Faris' arrival time in London = 3.30 pm + 3 h 54 mnt = 7.24 pm

3 0

3 years ago

Read 2 more answers

Lin and Diego both ran for 10 seconds, each at their own constant speed. Lin ran 40 meters and Diego ran 55 meters. 1. Who was m

Igoryamba

Answer:

Diego ran faster

Step-by-step explanation:

In the ten seconds, Diego ran 15 meters faster than Lin.

55 - 40 = 15

5 0

3 years ago