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saveliy_v [14]
3 years ago
14

,Line segment PQ is shown on a coordinate grid:

Mathematics
2 answers:
Helga [31]3 years ago
8 0

Answer:B

Step-by-step explanation:

Marina86 [1]3 years ago
6 0

Answer:

P'Q' is equal in length to PQ.

Step-by-step explanation:

Before rotation

P(-5, 3)

Q(-1, 3)

we get the length

L = √((-1-(-5))²+(3-3)²) = √((-4)²+(0)²) = 4

After rotation

P'(3, 5)

Q'(3, 1)

we get the length

L' = √((3-3)²+(1-5)²) = √((0)²+(-4)²) = 4

we can say that   L = L' = 4

P'Q' is equal in length to PQ.

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s344n2d4d5 [400]
V^2/(1-v^2/c^2)=R
v^2=R(1-v^2/c^2)
v^2=R-Rv^2/c^2
v^2-Rv^2/c^2=R
v^2(1-R/c^2)=R
v=sqrt(R/(1-R/c^2))
where R was original right side, dont forget plus minus
8 0
3 years ago
Decide whether ▱ABCD with vertices A(−2,3), B(2,3), C(2,−1), and D(−2,−1) is a rectangle, a rhombus, or a square. Select all nam
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D and B

Step-by-step explanation:

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3 years ago
1.6 × 10^6 ÷ 4.8 × 10^5. Give your answer as a mixed number​
Tasya [4]

Answer:

(10^11)/3

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6 0
2 years ago
High school math. Please answer everything in picture. Thank you................................................................
vaieri [72.5K]

Answer:

3) Cubic polynomial with four terms.

4) Linear polynomial with two terms.

5) The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618

The y-intercept is y=1

Please, see the attached graph.

6) The zeros are: -1.272, 1.272

The y-intercept is y=1

Please, see the attached graph.

7) Please, see the attached files.

8) Please, see the attached files.

9) Please, see the attached files.

10) Please, see the attached files.


Step-by-step explanation:

3) Degree is the maximun exponent that the variable "x" has, in this case 3, then this is a cubic polynomial, and it has four terms (6x^3, -7x^2, -10x, and -8).


4) Degree is the maximun exponent that the variable "x" has, in this case 1 (x=x^1), then this is a linear polynomial, and it has two terms (-10x, and 10).


5) f(x)=x^3-2x^2+1

Zeros:

f(x)=0→x^3-2x^2+1=0

Factoring:

    x^3-2x^2+0x+1  

 ! 1      -2       0    1

<u>1 !____1____-1_-1</u>

   1      -1        -1   0

   1x^2-1x      -1 = x^2-x-1

x^3-2x^2+1=0→(x-1)(x^2-x-1)=0

The zeros are:

x-1=0→x-1+1=0+1→x1=1

x^2-x-1=0

ax^2+bx+c=0; a=1, b=-1, c=-1

x=(-b+-sqrt(b^2-4ac))/(2a)

x=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))

x=(1+-sqrt(1+4))/2

x2=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618

x=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618

x3=(1+sqrt(5))/2

The zeros are: (1-sqrt(5))/2=-0.618, 1, and (1+sqrt(5))/2=1.618  

The y-intercept is:

x=0→f(0)=(0)^3-2(0)^2+1→f(0)=0-2(0)+1→f(0)=0-0+1→f(0)=1

The y-intercept is y=1.


6) f(x)=-x^4+x^2+1

Zeros:

f(x)=0→-x^4+x^2+1=0

Factoring: Multiplyng both sides of the equation by -1:

(-1)(-x^4+x^2+1)=(-1)(0)

x^4-x^2-1=0

(x^2)^2-(x^2)-1=0

Changing x^2 by t:

t^2-t-1=0

at^2+bt+c=0; a=1, b=-1, c=-1

t=(-b+-sqrt(b^2-4ac))/(2a)

t=(-(-1)+-sqrt((-1)^2-4(1)(-1))/(2(1))

t=(1+-sqrt(1+4))/2

t1=(1+-sqrt(5))/2=(1+2.236067978)/2=3.236067978/2=1.618033989=1.618

t2=(1-sqrt(5))/2=(1-2.236067978)/2=-1.236067978/2=-0.618033989=-0.618

t^2-t+1=0→(t-1.618)(t-(-0.618))=0→(t-1.618)(t+0.618)=0

and t=x^2, then:

(x^2-1.618)(x^2+0.618)=0

Factoring the first parentheses using a^2-b^2=(a+b)(a-b), with:

a^2=x^2→sqrt(a^2)=sqrt(x^2)→a=x

b^2=1.618→sqrt(b^2)=sqrt(1.618)→b=1.272

(x^2-1.618)(x^2+0.618)=0→(x+1.272)(x-1.272)(x^2+0.618)=0

The zeros are:

x+1.272=0→x+1.272-1.272=0-1.272→x1=-1.272

x-1.272=0→x-1.272+1.272=0+1.272→x2=1.272

The zeros are: -1.272, and 1.272  

The y-intercept is:

x=0→f(0)=-(0)^4+(0)^2+1→f(0)=-0+0+1→f(0)=1

The y-intercept is y=1.

3 0
2 years ago
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balu736 [363]
D) 1•x=5, 1•5=5 8•5=40, the answer is 40/5 CDs

E) 3/3=1, 26.25/3= 8.75. The answer is 1/8.75

F) 10/10=1, 120/10= 12. The answer is 1/12

G) 2•x=12, 2•6=12, 120•6= 720. The answer is 12/720

Hopefully this helped
5 0
2 years ago
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