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3 years ago

12

∠WXY and ∠ZXY are adjacent angles. What equation must always be TRUE? m∠WXY=m∠ZXY

1 answer:

Hitman42 [59]3 years ago

3 0

Answer:

m∠WXY+m∠ZXY=m∠WXZ

Step-by-step explanation:

You don't know the angles to be congruent (equal measures), complementary (measures add to 90 deg), or supplementary (measures add to 180 deg); all you know is that they are adjacent, so the sum of the measures of the two smaller angles equals the measure of the larger outer angle.

Answer: m∠WXY+m∠ZXY=m∠WXZ

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Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel

timama [110]

Answer:

$\displaystyle A=\frac{1}{192}$

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

$P=4x+3y$

We know the total fencing is 1/2 miles long, thus

$\displaystyle 4x+3y=\frac{1}{2}$

Solving for x

$\displaystyle x=\frac{\frac{1}{2}-3y}{4}$

The total area of the pasture is

$A=x.y$

Substituting x

$\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y$

$\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}$

Differentiating with respect to y

$\displaystyle A'=\frac{\frac{1}{2}-6y}{4}$

Equate to 0

$\displaystyle \frac{\frac{1}{2}-6y}{4}=0$

Solving for y

$\displaystyle y=\frac{1}{12}$

And also

$\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}$

Compute the second derivative

$\displaystyle A''=-\frac{3}{2}.$

Since it's always negative, the point is a maximum

Thus, the maximum area is

$\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}$

6 0

3 years ago

Which of the following is true of the location of the terminal side of an angle, Theta, whose sine value is One-half?

FinnZ [79.3K]

Answer:

Theta has a reference angle of 30° and is in Quadrant I or II

Step-by-step explanation:

Sin(theta) = ½

Basic angle: 30

Angles:

30,

180-30 = 150

Because sin is positive in quadrants 1 and 2

6 0

3 years ago

Read 2 more answers

Fractions and integer operations practice, I need help please.

olga2289 [7]

40/11 = 3.63
hope this helps

3 0

3 years ago

I need help on a question

schepotkina [342]

Given the figure of a regular pyramid

The base of the pyramid is a hexagon with a side length = 6

The lateral area is 6 times the area of one of the side triangles

So, the side triangle has a base = 6

The height will be:

$\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}$

so, the lateral area =

$6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}$

3 0

1 year ago

To make the expression 4x-2 + (2x + 4) equivalent to 2(6x + 7), what must be the value in the box?

podryga [215]

$4x - 2 + (2x + 4) = 2(6x + 7) \\$

$4x - 2 + 2x + 4 = 12x + 14$

Simplification

$6x + 2 = 12x + 14$

Subtract sides 2

$6x + 2 - 2 = 12x + 14 - 2$

Simplification

$6x = 12x + 12$

Subtract sides -12x

$6x - 12x = 12x - 12x + 12$

Simplification

$- 6x = 12$

Divided sides by -6

$\frac{ - 6}{ - 6}x = \frac{12}{ - 6} \\$

$x = - 2$

It is must be in the box buddy.

And we're done...♥️♥️♥️♥️♥️

5 0

3 years ago