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2 years ago

Help me PLEASEEEEEEEEEEEEE

Mathematics

1 answer:

maxonik [38]2 years ago

6 0

Answer:

D. 144

Step-by-step explanation:

Find volume of shipping container:

2 2/3 = 8/3

2 x 1 = 2

8/3 x 2/1 = 16/3

Find volume of cube:

1/3 x 1/3 = 1/9 x 1/3 = 1/27

Find how many boxes fit in container:

16/3 ÷ 1/27

16/3 x 27/1

Simplify:

16/1 x 9/1 = 144/1

Answer: 144

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Solve For Each Variable.

bixtya [17]

The first equation given $6=1-2n+5$

We have to add 1 and 5 to the right side first. We will get,

$6= 6 - 2n$

To get rid of 6 from the right side we have to subtract 6 from both sides.

$6-6 = 6-6-2n$

$6-6 = -2n$

$0=-2n$

To find n we have to move -2 to the other side by dividing both side by -2.

$0/-2 = -2n/-2$

$0= n$

$n=0$

So we have got the required answer for the first question.

The solution is n = 0.

The second equation given,

$8x-2 = -9+7x$

First we have to move 7x to the left side by subtracting it from both sides.

$8x-7x-2 = -9+7x-7x$

$8x-7x-2 = -9$

$x-2 = -9$

Now we have to move -2 to the right side by adding 2 to both sides.

$x-2+2 = -9+2$

$x = -9+2$

$x = -7$

We have got the required answer for the second question.

The solution is x = -7.

The third equation given,

$-8 = -(x+4)$

We have to get rid of that negative sign from both sides. As we have negative sign to both sides we can cancel it out. We will get,

$8=x+4$

Now we have to move 4 to left side by subtracting it from both sides.

$8-4 = x+4-4$

$8-4 = x$

$4=x$

$x = 4$

So we have got the required answer .

The solution is x = 4.

3 0

3 years ago

What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE

Nutka1998 [239]

Answer:

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),$ $\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

Step-by-step explanation:

Let points D, E and F have coordinates $(x_D,y_D),\ (x_E,y_E)$ and $(x_F,y_F).$

1. Midpoint M of segment DF has coordinates

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).$

2. Midpoint N of segment EF has coordinates

$\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0

3 years ago

Help on questions 11 and 15 Add and simplify if possible

Svetllana [295]

$\frac{16+x}{x^3}+\frac{7-4x}{x^3}=\frac{16+x+7-4x}{x^3}=\frac{23-3x}{x^3}\\\\======================================\\\\\frac{5}{t-1}+\frac{3}{t}=\frac{5t}{t(t-1)}+\frac{3(t-1)}{t(t-1)}=\frac{5t+3t-3}{t^2-t}=\frac{8t-3}{t^2-t}$

7 0

2 years ago

Read 2 more answers

A school P is 16km due west of a school Q. what is the bearing of Q from P

Irina-Kira [14]

Answer:

16Km due east of school P

Step-by-step explanation:

Given

A school P is 16km due west of a school Q

Thus, we can say that distance PQ = 16 km.

________________________________

now we have to find the bearing of Q from P

As distance is same

distance PQ = distance QP

Thus,

Distance will remain same of 16 km.

For direction,

If Q is west of P, then P will be east of Q $P------------------>Q$

as shown in figure P is west of Q,

now from point P , Q is west P.

Thus,

Bearing of School Q from P is 16Km due east of school P

7 0

3 years ago

Question 13 of 20 :

grin007 [14]

Answer:

C. 70.4

Step-by-step explanation:

22% × 320 =

(22 ÷ 100) × 320 =

(22 × 320) ÷ 100 =

7,040 ÷ 100 =

70.4

8 0

3 years ago