All categories

3 years ago

WHAT IS 18 TONS 42 POUNDS DEVIDED BY 3

Mathematics

2 answers:

Alecsey [184]3 years ago

4 0

A ton is 2,000 pounds so multiply 2,000 x 18

2,000 x 18 = 36,000 pounds, now add the 42 pounds to the 36,000

36,000 + 42 = 36,042 pounds. Now divide 36,042 by 3

36,042 ÷ 3 = 12,014

Your answer is 12,014 pounds

If you need your answer in tons it's 6.007 tons

Calculations for pounds to tons

12,014 ÷ 2,000 = 6.007

Hope this helps. :)

neonofarm [45]3 years ago

3 0

Just divide each value individually.

$\frac{18t}3 = 6t \\\\ \frac{42\ lbs}3 = 14\ lbs$

Thus a third of 18t. and 42 lbs. is 6t. and 14 lbs.

You might be interested in

Please help im really dumb also how do you delete other peoples post on this app

PIT_PIT [208]

It would be 15.60 ,correct me if am wrong :)

3 0

3 years ago

Read 2 more answers

How many times larger is 7400 than 74?

ladessa [460]

7400 / 74 = 100.....so 7400 is 100 times larger then 74

6 0

4 years ago

Read 2 more answers

Please guys help me out with this

Kisachek [45]

When you have a negative exponent, you move the variable with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

$x^{-2}$ or $\frac{x^{-2}}{1} =\frac{1}{x^2}$

$\frac{1}{2y^{-3}} =\frac{y^3}{2}$ (you don't move the "2" because "2" has an exponent of 1, only "y" has the negative exponent)

$\frac{5^{-2}}{y}=\frac{1}{(5^2)y} =\frac{1}{25y}$

When you multiply a variable with an exponent by a variable with an exponent, you add the exponents together. (But you can only combine the exponents when the variables are the same)

$x^2(y^3)=x^2y^3$ (You can't combine them because they have different variables of x and y)

$x^3(x^5)=x^{(3+5)}=x^8$

$y^3(y^2)=y^{(3+2)}=y^5$

There's two ways you can do this. Either by first making all the exponents positive then do subtraction, or multiply them then make all the exponents positive. I'm doing the 2nd way.

$2w^7u^{-2}w^{-6}*9y^{-6}*2y^9u$ You can combine the numbers 2 x 9 x 2 = 36

$36w^7u^{-2}w^{-6}y^{-6}y^9u$ Now multiply the terms with the same variables

$36(w^{7+(-6)})(u^{(-2+1)})(y^{(-6+9)})$

$36(w^1)(u^{-1})(y^3)$ Now make all the exponents positive

$\frac{36wy^3}{u}$

If you were to do subtraction, here is what you need to know:

When you divide a variable with an exponent by a variable with an exponent, you subtract the exponents together. (Reminder: they have to be the same variable in order to combine the exponents)

For example:

$\frac{x^3}{x^2} =x^{(3-2)}=x^1$ or x

$\frac{y^2}{y^4} =y^{(2-4)}=y^{-2}=\frac{1}{y^2}$

7 0

3 years ago

Find all the missing sides or angles in each right triangles

astra-53 [7]

In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle △XYZ is cut by AB¯¯¯¯¯¯¯¯ where A and B are midpoints of sides XZ¯¯¯¯¯¯¯¯ and YZ¯¯¯¯¯¯¯ respectively. AB¯¯¯¯¯¯¯¯ is called a midsegment of △XYZ. Note that △XYZ has other midsegments in addition to AB¯¯¯¯¯¯¯¯. Can you see where they are in the figure above?

If we construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and construct CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ respectively, we have the following figure and see that segments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ are midsegments of △XYZ.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

parallel to the third side. half as long as the third side.

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle △XYZ. Construct the midpoint A of side XZ¯¯¯¯¯¯¯¯.

By the Parallel Postulate, there is exactly one line though A that is parallel to side XY¯¯¯¯¯¯¯¯. Let’s say that it intersects side YZ¯¯¯¯¯¯¯ at point B. We will show that B must be the midpoint of XY¯¯¯¯¯¯¯¯ and then we can conclude that AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is parallel to XY¯¯¯¯¯¯¯¯.

We must show that the line through A and parallel to side XY¯¯¯¯¯¯¯¯ will intersect side YZ¯¯¯¯¯¯¯ at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side YZ¯¯¯¯¯¯¯.

Since XA¯¯¯¯¯¯¯¯≅AZ¯¯¯¯¯¯¯, we have BZ¯¯¯¯¯¯¯≅BY¯¯¯¯¯¯¯¯. Hence, by the definition of midpoint, point B is the midpoint of side YZ¯¯¯¯¯¯¯. AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is also parallel to XY¯¯¯¯¯¯¯¯.

Proof of 2. We must show that AB=12XY.

In △XYZ, construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and midsegments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ as follows:

First note that CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ by part one of the theorem. Since CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯∥XY¯¯¯¯¯¯¯¯, then ∠XAC≅∠BCA and ∠CAB≅∠ACX since alternate interior angles are congruent. In addition, AC¯¯¯¯¯¯¯¯≅CA¯¯¯¯¯¯¯¯.

Hence, △AXC≅△CBA by The ASA Congruence Postulate. AB¯¯¯¯¯¯¯¯≅XC¯¯¯¯¯¯¯¯ since corresponding parts of congruent triangles are congruent. Since C is the midpoint of XY¯¯¯¯¯¯¯¯, we have XC=CY and XY=XC+CY=XC+XC=2AB by segment addition and substitution.

So, 2AB=XY and AB=12XY. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

A. MN. B. The perimeter of the triangle △XYZ. A. Since O is a midpoint, we have XO=5 and XY=10. By the theorem, we must have MN=5. B. By the Midsegment Theorem, OM=3 implies that ZY=6; similarly, XZ=8, and XY=10. Hence, the perimeter is 6+8+10=24.

We can also examine triangles where one or more of the sides are unknown.

Example 2

Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment XY¯¯¯¯¯¯¯¯.

By the Midsegment Theorem we have 2x−6=12(18). Solving for x, we have x=152.

Lesson Summary

8 0

3 years ago

Albert and John are partners in a Bakery store. They needed $70,000 to start the business. They

viktelen [127]

$30,000 you’re welcome

3 0

4 years ago

Read 2 more answers