What is the Answer for X/3+11=16

Is y=1/2x-2 a function

zepelin [54]

Yes i think it is a function

8 0

3 years ago

Derivative using direct definition of derivative 2-x/2+x

Daniel [21]

Let f(x) = (2 - x)/(2 + x). By definition of the derivative,

$\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$

$\displaystyle f'(x)=\lim_{h\to0}\frac{\frac{2-(x+h)}{2+(x+h)}-\frac{2-x}{2+x}}h$

$\displaystyle f'(x)=\lim_{h\to0}\frac{\frac{(2-x-h)(2+x)-(2-x)(2+x+h)}{(2+x+h)(2+x)}}h$

$\displaystyle f'(x)=\lim_{h\to0}\frac{-4h}{h(2+x+h)(2+x)}$

$\displaystyle f'(x)=-4\lim_{h\to0}\frac1{(2+x+h)(2+x)}=\boxed{-\frac4{(2+x)^2}}$

3 0

2 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

What is the answer to number 4

laiz [17]

Y=-2x^2+2x-4. Should be correct

3 0

3 years ago

The area of a rectangle has decreased by 2%.

MissTica

Answer:

The other side was decreased to approximately .89 times its original size, meaning it was reduced by approximately 11%

Step-by-step explanation:

We can start with the basic equation for the area of a rectangle:

l × w = a

And now express the changes described above as an equation, using "p" as the amount that the width is changed:

(l × 1.1) × (w × p) = a × .98

Now let's rearrange both of those equations to solve for a / l. Starting with the first and easiest:

w = a/l

now the second one:

1.1l × wp = 0.98a

wp = 0.98a / 1.1l

1.1 wp / 0.98 = a/l

Now with both of those equalling a/l, we can equate them:

1.1 wp / 0.98 = w

We can then divide both sides by w, eliminating it

1.1wp / 0.98w = w/w

1.1p / 0.98 = 1

And solve for p

1.1p = 0.98

p = 0.98 / 1.1

p ≈ 0.89

So the width is scaled by approximately 89%

We can double check that too. Let's multiply that by the scaled length and see if we get the two percent decrease:

.89 × 1.1 = 0.979

That should be 0.98, and we're close enough. That difference of 1/1000 is due to rounding the 0.98 / 1.1 to .89. The actual result of that fraction is 0.89090909... if we multiply that by 1.1, we get exactly .98.

6 0

2 years ago