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Helen [10]
3 years ago
9

Mrs. Roberson has 5 more students in her class than Mr. Monroe. Together, they have 57 students. Write an equation that can be u

sed to find the number of students in Mr. Monroe’s (m) class. Solve the equation.
Mathematics
1 answer:
Ivahew [28]3 years ago
8 0

Answer:

Step-by-step explanation:

the basic number will be represented by X so, the equation will be

Mr. Monroe Mrs. Roberson

X X+5

57 = X + (X+5)

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Two boards are 26 meters long altogether. One board is 8 meters long. How long is the other board?
Aleonysh [2.5K]
I think the answer is 18 because together, they are 26 meters and one is 8 so u have to subtract the total with 8 which is 26-8. the answer to that would be 18.
ANSWER: 18 meters
6 0
3 years ago
Read 2 more answers
Rob laid three sticks end to end. The lengths of the sticks were 45 millimeters, 32 centimeters, and 4 decimeters. How long, in
Oksanka [162]

Answer:

<u>0.765 meters</u> long were the sticks when laid end to end.

Step-by-step explanation:

Given:

Rob laid three sticks end to end. The lengths of the sticks were 45 millimeters, 32 centimeters, and 4 decimeters.

Now, to find the length in meters of the sticks when laid end to end.

The lengths of the sticks are:

1st stick - 45 millimeters.

2nd stick - 32 centimeters.

3rd stick - 4 decimeters.

So, we convert the units of sticks into meters by using conversion factor:

1st stick:

1 millimeter = 0.001 meter.

45 millimeters = 0.001 × 45

45 millimeters = 0.045 meter.

2nd stick:

1 centimeter = 0.01 meter.

32 centimeters = 0.01 × 32

32 centimeters = 0.32 meter.

3rd stick:

1 decimeter = 0.1 meter.

4 decimeters = 0.1 × 4

4 decimeters = 0.4 meter.

Now, adding all the three sticks length to get the the length in meters of the sticks when laid end to end:

0.045+0.32+0.4\\\\=0.765\ meters.

Therefore, 0.765 meters long were the sticks when laid end to end.

5 0
4 years ago
7,000 is ten times as much as what number
Nezavi [6.7K]
700. To find out what number a value is ten times as much as, divide by 10.
5 0
4 years ago
A closed can, in a shape of a circular, is to contain 500cm^3 of liquid when full. The cylinder, radius r cm and height h cm, is
Gemiola [76]
To express the height as a function of the volume and the radius, we are going to use the volume formula for a cylinder: V= \pi r^2h
where
V is the volume 
r is the radius 
h is the height 

We know for our problem that the cylindrical can is to contain 500cm^3 when full, so the volume of our cylinder is 500cm^3. In other words: V=500cm^3. We also know that the radius is r cm and height is h cm, so r=rcm and h=hcm. Lets replace the values in our formula:
V= \pi r^2h
500cm^3= \pi (rcm^2)(hcm)
500cm^3=h \pi r^2cm^3
h= \frac{500cm^3}{ \pi r^2cm^3}
h= \frac{500}{ \pi r^2}

Next, we are going to use the formula for the area of a cylinder: A=2 \pi rh+2 \pi r^2
where
A is the area 
r is the radius 
h is the height

We know from our previous calculation that h= \frac{500}{ \pi r^2}, so lets replace that value in our area formula:
A=2 \pi rh+2 \pi r^2
A=2 \pi r(\frac{500}{ \pi r^2})+2 \pi r^2
A= \frac{1000}{r} +2 \pi r^2
By the commutative property of addition, we can conclude that:
A=2 \pi r^2+\frac{1000}{r}
7 0
4 years ago
1. Write a two-column proof.
Alexandra [31]

Answer:

Check it below, please

Step-by-step explanation:

Hi there!

Let's prove segment AB is perpendicular to CD. Attention to the fact that a two column proof has to be concise. So all the comments can't be exhaustive, but as short as possible.

Let's recap: An isosceles triangle is one triangle with at least 2 congruent angles.

Statement                     Reason

\overline{AB} \cong \overline{AC}\\                     Given

\widehat{ACB} \:bisects\: \overline{AB}        Isosceles Triangle the altitude, the bisector coincide.          

\overline{AD} \cong \overline{DB}               Bisector equally divide a line segment into two congruent      

m\angle CDA =m \angle CBD=90^{\circ} Right angles, perpendicular lines.                  

\overline{CD}\perp \overline{AB}                    Perpendicular Line segment      

8 0
3 years ago
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