All categories

3 years ago

Please help i need help geometry is very hard for me i am dumb

Mathematics

1 answer:

stiks02 [169]3 years ago

3 0

Answer:

G. ABD = 74

H. DBC = 206

I. XYW = 33.75

J. WYZ = 46.25

Step-by-step explanation:

For G and H: You have a straight line (ABC) with another line coming off of it, creating two angles (ABD and DBC). A straight line has an angle of 180 degrees. This means that the two angles from the straight line when combined will give you 180 degrees. Solve for x.

ABD + DBC = ABC

(1/2x + 20) + (2x - 10) = 180

1/2x + 20 + 2x - 10 = 180

5/2x + 10 = 180

5/2x = 170

x = 108

Now that you have x, you can solve for each angle.

ABD = 1/2x + 20

ABD = 1/2(108) + 20

ABD = 54 + 20

ABD = 74

DBC = 2x - 10

DBC = 2(108) - 10

DBC = 216 - 10

DBC = 206

For I and J: For these problems, you use the same concept as before. You have a right angle (XYZ) that has within it two other angles (XYW and WYZ). A right angle has 90 degrees. Combine the two unknown angles and set it equal to the right angle. Solve for x.

XYW + WYZ = XYZ

(1 1/4x - 10) + (3/4x + 20) = 90

1 1/4x - 10 + 3/4x + 20 = 90

2x + 20 = 90

2x = 70

x = 35

Plug x into the angle values and solve.

XYW = 1 1/4x - 10

XYW = 1 1/4(35) - 10

XYW = 43.75 - 10

XYW = 33.75

WYZ = 3/4x + 20

WYZ = 3/4(35) + 20

WYZ = 26.25 + 20

WYZ = 46.25

You might be interested in

4x - 5 = 11 is an example of a

Tanya [424]

Answer:

MATH PROBLEM

Step-by-step explanation:

3 0

3 years ago

Please help me 15 points :)

maw [93]

Answer:I don't really know sorry but I think its the second one

Step-by-step explanation:

7 0

2 years ago

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

6 0

3 years ago

What does f(-1) =_ and x=_ ?

lubasha [3.4K]

Answer:

F(-1) = 1

x = 3

Step-by-step explanation:

1) Assume that f(-1) = y1

So that point A (x1 = -1; y1) is a point lying in the graph representing the equation y = f(x)

As it can be seen in the figure, the graph representing that equation crosses point (-1; 1)

=> Point A has y1 = 1

=> f (-1) =1

2) Assume that f(x2) = -2

So that point B (x2; y2 = -2) is a point on the graphy representing the equation y = f(x)

As indicated in the figure, the graph crosses point (3; -2)

=> Point B has x2 = 3

=> When f(x) = -2, x = 3

5 0

3 years ago

In the given figure p=120° and y : z=2 : 3 Find x, y and z

Sindrei [870]

Answer:

x = 60°

y = 72°

z = 48°

Step-by-step explanation:

x = 180 - p = 180 - 120 = 60

Since y;z = 3:2 , 2y = 3z or z = 2y/3

Now, x + y + z = 180

60 + y + 2y/3 = 180 Multiply thru by 3 to remove the fraction

180 + 3y + 2y = 540

180 + 5y = 540

5y = 360

y = 72

z = 2(72)/3 = 48

Check: 60 + 72 + 48 = 180

5 0

3 years ago