All categories

4 years ago

Angles

Mathematics

1 answer:

coldgirl [10]4 years ago

4 0

<h3>Answer: XWY and STR</h3>

I tend to think of parallel lines as train tracks (the metal rail part anyway). Inside the train tracks is the interior region, while outside the train tracks is the exterior region. Alternate exterior angles are found here. Specifically they are angles that are on opposite or alternate sides of the transversal cut.

Both pairs of alternate exterior angles are shown in the diagram below. They are color coded to help show how they pair up and which are congruent.

A thing to notice: choices B, C, and D all have point W as the vertex of the angles. This means that the angles somehow touch or are adjacent in some way due to this shared vertex point. However, alternate exterior angles never touch because parallel lines never do so either. We can rule out choices B,C,D from this reasoning alone. We cannot have both alternate exterior angles on the same exterior side of the train tracks. Both sides must be accounted for.

You might be interested in

need help please. Problem: Construct a triangle with interior angle measures of 45° and 45°. The third angle of the triangle is

REY [17]

The angles of a triangle add to 180 degrees so the remaining angle must be 90°.

This triangle is half a square. It's one of the two tired triangles of trig, the other being half an equilateral triangle, 30°/60°/90°.

4 0

4 years ago

Read 2 more answers

How me solve this please

marissa [1.9K]

Answer:

Step-by-step explanation:

100/25 = 4

but we have 5 students absent so 25-5=20

100/20 = 5

so A

5 0

2 years ago

Create 2 lines thatare parallel to 8x + 2y = 7?

Alinara [238K]

d1 = ax + by + c = 0 and d2= mx + ky + f = 0

if they are parallel :

$\frac{a}{m} = \frac{b}{k} \neq \frac{c}{f}$

8x + 2y = 7

8x + 2y - 7 = 0

8x + 2y -9 = 0
4x + y -6 = 0
16x + 4y - 19 = 0

you can create more whatever you want =)

Hope this helps ^-^

7 0

3 years ago

The data in the table represents the average number of daylight hours each month in Springfield in 2015, rounded to the nearest

Lera25 [3.4K]

To do part B, use the equation that you wrote for part A. We need to figure out what month number March of 2020 will be; we started in January 2015 at month 1; month 2 was Feb. 2015; month 3 was Mar. 2015; etc. to month 12 at Dec. 2015. Then month 13 will be Jan. 2016; month 13+12=25 will be Jan. 2017; month 25+12=37 will be Jan. 2018; month 37+12=49 will be Jan. 2019; month 49+12=61 will be Jan. 2020; month 62 will be Feb. 2020; and month 63 will be March 2020.

63 is the number you will substitute for x in the equation you've already written for part A.

8 0

3 years ago

Read 2 more answers

The Ace Novelty company produces two souvenirs: Type A and Type B. The number of Type A souvenirs, x, and the number of Type B s

Molodets [167]

Answer:

500 type A; 3500 type B

Step-by-step explanation:

The method of Lagrange multipliers can solve this quickly. For objective function f(x, y) and constraint function g(x, y)=0 we can set the partial derivatives of the Lagrangian to zero to find the values of the variables at the extreme of interest.

These functions are ...

$f(x,y)=4x+2y\\g(x,y)=2x^2+y-4$

The Lagrangian is ...

$\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda g(x,y)\\\\\text{and the partial derivatives are ...}\\\\\dfrac{\partial \mathcal{L}}{\partial x}=\dfrac{\partial f}{\partial x}+\lambda\dfrac{\partial g}{\partial x}=4+\lambda (4x)=0\ \implies\ x=\dfrac{-1}{\lambda}\\\\\dfrac{\partial \mathcal{L}}{\partial y}=\dfrac{\partial f}{\partial y}+\lambda\dfrac{\partial g}{\partial y}=2+\lambda (1)=0\ \implies\ \lambda=-2$

$\dfrac{\partial\mathcal{L}}{\partial\lambda}=\dfrac{\partial f}{\partial\lambda}+\lambda\dfrac{\partial g}{\partial\lambda}=0+2x^2+y-4=0\ \implies\ y=4-2x^2\\\\\text{We know $\lambda$, so we can find x and y:}\\\\x=\dfrac{-1}{-2}=0.5\\\\y=4-2\cdot 0.5^2=3.5$

Since x and y are in thousands, maximum profit is to be had when the company produces ...

500 Type A souvenirs, and 3500 Type B souvenirs

3 0

3 years ago