Please help me i beg of u due today

Mathematics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

51) D. a dilation by a scale factor of 1/2 with the origin as the center of dilation.

44) B. The angles measures of △ A'B'C' are the same as those of △ ABC, but the side lengths of △ A'B'C' are half the size of those of △ ABC

Step-by-step explanation:

51) option D is the only one that changes the shape of the figure, which is the only way it wont be congruent

44) angles don't change when the figure is rotated or dilated. the shape is half the size, so the sides are half as long.

i got you! idk why i'm doing this but i know if i was your age this would have made my day. you're only 15 so don't take life too seriously, take lots of risks, and always take the easy way out <3

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Need help with the blanks

Crazy boy [7]

Answers:
33. Angle R is 68 degrees
35. The fraction 21/2 or the decimal 10.5
36. Triangle ACG
37. Segment AB
38. The values are x = 6; y = 2
40. The value of x is x = 29
41. C) 108 degrees
42. The value of x is x = 70
43. The segment WY is 24 units long
------------------------------------------------------
Work Shown:
Problem 33)
RS = ST, means that the vertex angle is at angle S
Angle S = 44
Angle R = x, angle T = x are the base angles
R+S+T = 180
x+44+x = 180
2x+44 = 180
2x+44-44 = 180-44
2x = 136
2x/2 = 136/2
x = 68
So angle R is 68 degrees
-----------------
Problem 35)
Angle A = angle H
Angle B = angle I
Angle C = angle J
A = 97
B = 4x+4
C = J = 37
A+B+C = 180
97+4x+4+37 = 180
4x+138 = 180
4x+138-138 = 180-138
4x = 42
4x/4 = 42/4
x = 21/2
x = 10.5
-----------------
Problem 36)
GD is the median of triangle ACG. It stretches from the vertex G to point D. Point D is the midpoint of segment AC
-----------------
Problem 37)
Segment AB is an altitude of triangle ACG. It is perpendicular to line CG (extend out segment CG) and it goes through vertex A.
-----------------
Problem 38)
triangle LMN = triangle PQR
LM = PQ
MN = QR
LN = PR
Since LM = PQ, we can say 2x+3 = 5x-15. Let's solve for x
2x+3 = 5x-15
2x-5x = -15-3
-3x = -18
x = -18/(-3)
x = 6
Similarly, MN = QR, so 9 = 3y+3
Solve for y
9 = 3y+3
3y+3 = 9
3y+3-3 = 9-3
3y = 6
3y/3 = 6/3
y = 2
-----------------
Problem 40)
The remote interior angles (2x and 21) add up to the exterior angle (3x-8)
2x+21 = 3x-8
2x-3x = -8-21
-x = -29
x = 29
-----------------
Problem 41)
For any quadrilateral, the four angles always add to 360 degrees
J+K+L+M = 360
3x+45+2x+45 = 360
5x+90 = 360
5x+90-90 = 360-90
5x = 270
5x/5 = 270/5
x = 54
Use this to find L
L = 2x
L = 2*54
L = 108
-----------------
Problem 42)
The adjacent or consecutive angles are supplementary. They add to 180 degrees
K+N = 180
2x+40 = 180
2x+40-40 = 180-40
2x = 140
2x/2 = 140/2
x = 70
-----------------
Problem 43)
All sides of the rhombus are congruent, so WX = WZ.
Triangle WPZ is a right triangle (right angle at point P).
Use the pythagorean theorem to find PW
a^2+b^2 = c^2
(PW)^2+(PZ)^2 = (WZ)^2
(PW)^2+256 = 400
(PW)^2+256-256 = 400-256
(PW)^2 = 144
PW = sqrt(144)
PW = 12
WY = 2*PW
WY = 2*12
WY = 24

3 0

3 years ago

Number 4 Pleases help need now

levacccp [35]

I believe the answer is 23 :)

3 0

4 years ago

What is dh/dx if h(x) = cπ, where c is a constant? Please show work. Will mark the best answer as brainliest.

Svetllana [295]

Answer:

$h'(x)=0$

Step-by-step explanation:

By definition, the derivative of $f(x)$ is given by:

$\displaystyle \frac{dy}{dx}=f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Given $h(x)=c\pi$ , $h(x+4)$ must also be $c\pi$ . Therefore, we have:

$\displaystyle h'(x)=\lim_{h\rightarrow 0}\frac{c\pi - c\pi }{h},\\h'(x)=\lim_{h\rightarrow 0}\frac{0}{h}=\boxed{0}$

Note that the derivative of a constant is always zero since for $f(x)=c$ , where $c$ is some constant, $f(x)=f(x+h)$ and we obtain $c-c$ in the numerator, yielding a final answer of zero.