I need help pls due tomorrow pls I just need part B done plsss
1 answer:
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Answer: 
M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)
see graph below
Explanation:Given:
The image of a quadrilateral on a coordinate plane
To find:
The coordinates of the new image after dilation of 1/2 have been applied to the original image.
Then graph the coordinates
First, we need to state the coordinates of the original image:
M = (3, -2)
F = (4, -2)
L = (1, -5)
W = (5, -5)
Next, we will apply a scale factor of 1/2:
The new coordinates:
M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)
M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)
Plotting the coordinates:
Answer:
B and D
Step-by-step explanation:
The quadratic formula is . Here a=20, b=-26, and c=8.
Substitute and you'll have:
This means the solutions are 26 - 6/ 40 and 26 + 6/ 40.
It simplifies to 1/2 and 4/5.
Answers:
A = 120
b = 45.0
c = 33.2
Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.
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Explanation:
We are given the following information
B = 35
C = 25
a = 68
We need to find the following
A, b, c
where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.
-----------------------
First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120
Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'
-----------------------
We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth
-----------------------
Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth
A = 120
b = 45.0
c = 33.2
Side Note: only one triangle is possible
See attached for a visual. I used GeoGebra to draw the triangle.
-------------------------------------------------------------
-------------------------------------------------------------
Explanation:
We are given the following information
B = 35
C = 25
a = 68
We need to find the following
A, b, c
where the lower case letters represent the side lengths; the upper case letters are the angles. The angles are opposite their corresponding sides. For instance, side lowercase b is opposite angle uppercase B. The other letters are positioned the same way.
-----------------------
First use the idea that for any triangle, the three angles (A,B,C) must add to 180 degrees
A+B+C = 180
Replace B and C with 35 and 25. Solve for angle A
A+35+25 = 180
A+60 = 180
A+60-60 = 180-60
A = 120
Now that we know the three angles A = 120, B = 35, C = 25, we can find the missing sides 'b' and 'c'
-----------------------
We will use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(120)/68 = sin(35)/b
b*sin(120) = 68*sin(35) <<--- cross multiply
b = 68*sin(35)/sin(120)
b = 45.037013350222 <<--- use a calculator; this value is approximate
b = 45.0 <<--- round to the nearest tenth
-----------------------
Do the same for side c
sin(A)/a = sin(C)/c
sin(120)/68 = sin(25)/c
c*sin(120) = 68*sin(25) <<--- cross multiply
c = 68*sin(25)/sin(120)
c = 33.1838323365404 <<--- use a calculator; this value is approximate
c = 33.2 <<--- round to the nearest tenth