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bulgar [2K]
3 years ago
12

What is the volume of 7cm 4cm 2cm

Mathematics
2 answers:
lubasha [3.4K]3 years ago
6 0
Answer: 56cm^3

To find the Volume of a shape we do…
Base * Height * Length

In this case these value will be
7cm * 4cm * 2cm = 56cm^3
Viktor [21]3 years ago
5 0

Answer:

7 x 2 x 3 = 42 cm3

Step-by-step explanation:

i hope its right

stan got7

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Find b: If a = 9 and c = 15<br> С<br> a<br> b
pogonyaev

Answer:

b=12

Step-by-step explanation:

a^2+b^2=c^2

(9×9) + b^2= 15×15

81+ b^2= 225

-81. -81

b^2=144

√. √

b= 12

3 0
3 years ago
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In City A, the temperature rises 7 degrees F from 8AM to 9AM. Then the temperature drops 6 degrees F from 9AM to 10AM. In City B
madam [21]

Answer:

is there like a image of this eqation?

Step-by-step explanation:

5 0
3 years ago
A venue plans to host trucking events regularly throughout the year. The expected revenue from each event is $100,000. The cost
iris [78.8K]

Answer:

The venue has to host 47 events throughout the year

Step-by-step explanation:

Profit:

Profit is revenue subtracted by costs.

Per event:

$100,000 revenue

$50,000 costs

$100,000 - $50,000 = $50,000 profit.

If the venue wants to make a total profit margin of $2,350,000, how many events does the venue have to host throughout the year?

1 event - $50,000 profit.

x events - $2,350,000 profit.

50000x = 2350000

x = \frac{2350000}{50000}

x = 47

The venue has to host 47 events throughout the year

6 0
3 years ago
Does anyone know how to do simple interest for 7th grade math? I need a tutorial or an example please, thanks. A question exampl
lions [1.4K]

Answer:

750/100 X 7 = 52.50

52.50 X 3 = $157.50

Step-by-step explanation:

divide principal amount by 100 then multiply by interest percentage. this gave 52.50

now multiply the interest amount by number of years

in this case 3

52.50 X 3 = $157.50

3 0
3 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
MatroZZZ [7]

Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.

Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

Answer: the mean weight of the 11-person team is 208\dfrac{10}{11}\ lb.

4 0
3 years ago
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