What is the answer to 56/-8?

A cube has a volume of 7 to the 3rd power cubic centimeters . what is its surface area?

cricket20 [7]

This cube has 6 faces, and all side lengths are equal.

The volume of a cube with sides of length L is given by L^3.

The problem says that the volume is 7^3, which means that each side is 7

The surface area of a cube of length L is 6*L^2. Since L = 7, the surface area is:

6*7^2 = 6*49 = 294 cm^2

8 0

2 years ago

8,508,400 divided by 10 to the sixth power

irga5000 [103]

Your answer for your problem

4 0

2 years ago

Marianne has been collecting donations for her biscuit stall at the school summer fayre. There are some luxury gift tins of bisc

LenaWriter [7]

Answer:

Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80

Solution:

Number of luxury gift tins of biscuits sold at £5 each: x
Amount received by the number of luxury gift tins of biscuits sold: £5 x Number of normal packets sold at £1 each: y
Amount received by the number of normal packets sold: £1 y
Number of mini- packs of 2 biscuits sold at 10p each: z
Amount received by the number of 2 biscuists sold: 10p z =£0.1z

She tells Amy that she has received exactly 100 donations in total:
(1) x+y+z=100

A collective value of £100:
Total amount receibed: £5 x + £1 y+ £0.1z= £100 → £ (5x+1y+0.1z)= £100 → 5x+y+0.1z=100 (2)

Her stock of £1 packets is very low compared with the other items:
y<<x
y<<z
Then we can despice the value of "y" and solve for "x" and "z":
y=0:
(1) x+y+z=100→x+0+z=100→x+z=100 (3)
(2) 5x+y+0.1z=100→5x+0+0.1z=100→5x+0.1z=100 (4)

Solving the system of equations (3) and (4) using the substitution method:
Isolating z in equation (3): Subtrating x from both sides of the equation:
(3) x+z-z=100-x→z=100-x

Replacing z by 100-x in the equation (4):
(4) 5x+0.1z=100→5x+0.1(100-x)=100
Eliminating the parentheses:
5x+0.1(100)-0.1x=100→5x+10-0.1x=100
Adding similar terms on the left side of the equation:
4.9x+10=100
Solving for x: Subtrating 10 from both sides of the equation:
4.9x+10-10=100-10→4.9x=90
Dividing both sides of the equation by 4.9:
4.9x/4.9=90/4.9→x=18.36734693

Replacing x=18.36734693 in the equation:
z=100-x→z=100-18.36734693→z=81.63265306

We can round x and z: x=18 and z=81 and solve for y in equations (1) and (2):
(1) x+y+z=100→18+y+81=100
Adding similar terms on the left side of the equation:
99+y=100
Subtracting 99 from both sides of the equation:
99+y-99=100-99→y=1

(2) 5x+y+0.1z=100→5(18)+y+0.1(81)=100→90+y+8.1=100→98.1+y=100→98.1+y-98.1=100-98.1→y=1.9→y=2 different to 1

Suppose y=1. Replacing y=2 in equations (1) and (2):
(1) x+y+z=100→x+1+z=100→x+1+z-1=100-1→x+z=99
(2) 5x+y+0.1z=100→5x+1+0.1z=100→5x+1+0.1z-1=100-1→5x+0.1z=99

Repeating the process:
z=99-x
5x+0.1z=99→5x+0.1(99-x)=99→5x+9.9-0.1x=99→4.9x+9.9=99→4.9x+9.9-9.9=99-9.9→4.9x=89.1→4.9x/4.9=89.1/4.9→x=18.18367346
z=99-18.18367346→z=80.81632653
x=18 and z=80
(1) x+y+z=100→18+y+80=100→y+98=100→y+98-98=100-98→y=2
(2) 5x+y+0.1z=100→5(18)+y+0.1(80)=100→90+y+8=100→y+98=100→y+98-98=100-98→y=2

Answer:
Number of luxury gift tins of biscuits sold at £5 each: 18
Number of normal packets sold at £1 each: 2
Number of mini- packs of 2 biscuits sold at 10p each: 80

5 0

3 years ago

Read 2 more answers

Marty and Stewart are stuffing envelopes with index cards. They are putting x index cards in each envelope. When they are finish

Yanka [14]

Abac then in the other century there was a lot of killings

Step-by-step explanation:

4 0

2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

IRINA_888 [86]

Answer:

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = 0.02145 = 2.1 %

Step-by-step explanation:

Explanation:-

Given mean of the Population 'μ ' = 51 months

Standard deviation of the Population 'σ' = 11 months

Let 'X' be the random variable of Normal distribution

Let 'X' = 73

$Z = \frac{x-mean}{S.D} = \frac{73-51}{11} = 2$

Let 'X' = 84

$Z = \frac{84-51}{11} = 3$

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = P( 2 < Z < 3)

= P( Z<3) - P( Z <2)

= 0.5 + A(3) - ( 0.5 + A(2))

= A(3) - A( 2)

= 0.49865 - 0.4772 ( From Normal table)

= 0.02145

P( 73 < x < 84 ) = 0.02145

The approximate percentage of cars that remain in service between 73 and 84 months

P( 73 < x < 84 ) = 0.02145 = 2.1 %

7 0

3 years ago