Please tell i really need the answer:

Please help me with this guys

coldgirl [10]

Answer:

reflection across the x axis then reflection across the y axis

Step-by-step explanation:

5 0

3 years ago

How do you write 1.8 × 10^1 in standard form?

gayaneshka [121]

Answer:that would be 18

Step-by-step explanation:

7 0

3 years ago

Plz help me get the right answer to these

Citrus2011 [14]

- $b^{4}$ + 6 $b^{3}$

divide each term on the numerator by 7b

= - 7 $b^{5}$ / 7b + 42 $b^{4}$ / 7b

= - $b^{4}$ + 6 $b^{3}$

4 0

3 years ago

Read 2 more answers

PLEASE ANSWER ASAP! SHOW ALL WORK! WILL GIVE BRAINLIEST!

Iteru [2.4K]

1) $x^2=x\cdot x$ , and $x(x+3)=x^2+3x$ . Subtracting this from the numerator gives a remainder of

$(x^2-13x-48)-(x^2+3x)=-16x-48$

$-16x=-16\cdot x$ , and $-16(x+3)=-16x-48$ . Subtracting this from the previous remainder gives a new remainder of

$(-16x-48)-(-16x-48)=0$

This means that

$\dfrac{x^2-13x-48}{x+3}=x-16$

2) $3x^3=3x^2\cdot x$ , and $3x^2(x+2)=3x^3+6x^2$ . Subtracting this from the numerator gives a remainder of

$(3x^3-x^2-7x+6)-(3x^3+6x^2)=-7x^2-7x+6$

$-7x^2=-7x\cdot x$ , and $-7x(x+2)=-7x^2-14x$ . Subtracting this from the previous remainder gives a new remainder of

$(-7x^2-7x+6)-(-7x^2-14x)=7x+6$

$7x=7\cdot x$ , and $7(x+2)=7x+14$ . Subtracting this from the previous remainder gives a new remainder of

$(7x+6)-(7x+14)=-8$

This means that

$\dfrac{3x^3-x^2-7x+6}{x+2}=3x^2-7x+7-\dfrac8{x+2}$

3) $x+2$ will be a factor of $x^3+3x^2-10x-24$ if dividing the latter by $x+2$ leaves a remainder of 0.

$x^3=x^2\cdot x$ , and $x^2(x+2)=x^3+2x^2$ . Subtracting this from the numerator gives a remainder of

$(x^3+3x^2-10x-24)-(x^3+2x^2)=x^2-10x-24$

$x^2=x\cdot x$ , and $x(x+2)=x^2+2x$ . Subtracting this from the previous remainder gives a new remainder of

$(x^2-10x-24)-(x^2+2x)=-12x-24$

$-12x=-12\cdot x$ , and $-12(x+2)=-12x-24$ . Subtracting this from the previous remainder gives a new remainder of

$(-12x-24)-(-12x-24)=0$

and since the remainder is 0, $x+2$ is indeed a factor of $x^3+3x^2-10x-24$ .

5 0

3 years ago

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