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4 years ago

If a bag contains 10,500 ounces (oz) of sand, about how many pounds (lb) does it contain? [1 pound= 16 ounces]

Mathematics

2 answers:

madreJ [45]4 years ago

5 0

Answer: A: 656.25 lb

The bag contains 656.25 lb of sand.

Step-by-step explanation:

Given: A bag contains sand = 10,500 ounces

Also , 1 pound = 16 ounces.

Therefore to convert the given value to pounds, we need to divide the given value of ounces by 16, we get

$10500\text{ ounces}=\dfrac{10500}{16}\\\\\Rightarrow10500\text{ ounces}=656.25\text{pounds}$

Hence, the bag contains 656.25 lb of sand.

Ivahew [28]4 years ago

4 0

Okay so 1 pound = 16 ounces, and there is 10,500 ounces of sand.

Simply divide 10,500 by 16

10,500 / 16 = 656.25.

So your answer would be A.

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Triangle $PQR$ is isosceles and the measure of angle $R$ is $40^\circ$. The possible measures of angle $P$ are $x,y,z$. What is

Alisiya [41]

Answer:

The possible value for angle P will be 100

And sum will also 100 because it has only one value.

Step-by-step explanation:

Given:

angle R measures 40 degree and PQR is isosceles triangle.

To Find:

Measure angle P.

Solution:

The given triangle is isosceles hence it will have 2 same angle included in it.

And sum of all angles of a triangle is 180 degree.

let same angles be x and other be y

So ,

x+x+y=180.............. sum of all angles.

2x+y=180.

Now to find possible values for y ,with x=40,

2x+y=180

80+y=180

y=100

So being x= 40 y =100 which satisfy given angle R condition.

now let y=80,then x should satisfy 40 degree value, so

2x+80=180

2x=100

x=50

it doest not satisfy the given value hence possible for y will be 100.

Since x=40 degree.

8 0

4 years ago

Which is NOT true? A 7 + 7 = 19 - 5 B 9 + 9 = 11 + 7 C 19 = 13 + 5 D 20 - 12 = 4 + 4

Serjik [45]

Answer:

Step-by-step explanation:

13+5=18 not 19 therefore C is not true

8 0

3 years ago

Read 2 more answers

The endpoints of segment RS are R(-4, 3) and S(8, -5). Complete each statement using a fraction.

Ray Of Light [21]

Answer:

a. (-1 , 1) is the point at 1 : 3 of the way from R to S

b. (5 , -3) is the point at 3 : 1 of the way from R to S

Step-by-step explanation:

* Lets explain how to solve the problem

- If point (x , y) divide a line segment whose end points are

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ , at ratio

$m_{1}:m_{2}$ from the first point, then

$x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and

$y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

* Lets solve the problem

∵ RS is a line segment whose end points are R (-4 , 3) and S (8 , -5)

∵ Point (-1 , 1) divides it at ratio $m_{1}:m_{2}$ from R

- By using the rule above

∴ $-1=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}$

- By using cross multiplication

∴ $(-1)m_{1}+(-1)m_{2}$ = $-4m_{2}+8m_{1}$

- By collecting $m_{1}$ in one side and $m_{2}$ in the

other side

∴ $3m_{2}=9m_{1}$

- Divide both sides by 3

∴ $m_{2}=3m_{1}$

∴ $m_{1}=\frac{1}{3}m_{2}$

∴ (-1 , 1) is the point at 1 : 3 of the way from R to S

∵ RS is a line segment whose end points are R (-4 , 3) and S (8 , -5)

∵ Point (5 , -3) divides it at ratio $m_{1}:m_{2}$ from R

- By using the rule above

∴ $5=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}$

- By using cross multiplication

∴ $5m_{1}+5m_{2}$ = $-4m_{2}+8m_{1}$

- By collecting $m_{1}$ in one side and $m_{2}$ in the

other side

∴ $9m_{2}=3m_{1}$

- Divide both sides by 3

∴ $3m_{2}=m_{1}$

∴ $\frac{m_{1}}{m_{2}}=3$

∴ (5 , -3) is the point at 3 : 1 of the way from R to S

8 0

4 years ago

F(x) = -7x - 19 g(x) = 0.75x + 4.25 x=

Xelga [282]

Answer:x=-14.75

Step-by-step explanation:

f(x) = -7x - 19=-19

g(x) = 0.75x + 4.25 =4.25

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2 years ago

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lbvjy [14]

Answer:

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Step-by-step explanation:

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3 years ago