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igomit [66]
3 years ago
7

Please help I don’t know how to do this

Mathematics
1 answer:
mina [271]3 years ago
5 0

Answer:

f(x) = -2x + 7, when x > 1

f(x) = 4x + 1 when x ≤ 1

Step-by-step explanation:

The right side graph passe through points (1,5) and (2,3).  

Hence, the equation of the line  

\frac{y - 5}{5 - 3} = \frac{x - 1}{1 - 2}

⇒ (y - 5) = 2(1 - x)

⇒ y - 5 = 2 - 2x

⇒ y = -2x +7.......... (1)

Now, the left side graph passes through points (1,5) and (0,1)

Therefore, the equation of the graph will be  

\frac{y - 5}{5 - 1} = \frac{x - 1}{1 - 0}

⇒ y - 5 = 4(x - 1)

⇒ y = 4x + 1 ........ (2)

Hence, the equation (1) gives the function in the right side of the given graph which does not include the point (1,5) and extends to the right from this point.

Therefore, f(x) = -2x + 7, when x > 1 (Answer)

Again, equation (2) gives the function in the left side of the given graph which includes the point (1,5) and extends to the left from this point.

Therefore, f(x) = 4x + 1 when x ≤ 1 (Answer)

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VARVARA [1.3K]

Answer:

48

Step-by-step explanation:

u • w

= (i + 6j ) • (9i - 2j )

= (1 × 9 ) + (6 × - 2 )

= 9 + (- 12)

= 9 - 12

= - 3

v • w

= (5i - 3j ) • (9i - 2j )

= (5 × 9) + (- 3 × - 2)

= 45 + 6

= 51

Then

u • w + v • w

= - 3 + 51

= 48

7 0
3 years ago
I don’t understand this
Oxana [17]

You have to multiply whats outside the parenthesis with everything that is inside, so

x^{-3}y^0.(x^2-3x^5y^4)

x^{-3}y^0.x^2-x^{-3}y^0.3x^5y^4

Multiplication of same bases we sum the exponents

x^{-3+2}y^0-3x^{-3+5}y^{0+4}

x^{-1}.1-3x^2y^4

x^{-1}-3x^2y^4

Alternative B.

7 0
3 years ago
Helppppp plzzzz ASAP!!!!!!<br> Thank you!!!!!!
Zielflug [23.3K]

Answer:

option 4.

16 square units

Step-by-step explanation:

as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.

P = (2 , 4)

S = (4 , 2)

we have to subtract the values ​​of p from s

PS = (4 - 2  , 2 - 4)

PS = (2 , -2)

by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2

h: hypotenuse

c1: leg 1

c2: leg 2

PS^2 = 2^2 + -2^2

PS = √ 4 + 4

PS = √8

PS = 2√2

S = (4 , 2)

R = (8 , 6)

SR = (8-4  ,  6-2)

SR = (4 , 4)

by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2

h: hypotenuse

c1: leg 1

c2: leg 2

SR^2 = 4^2 + 4^2

SR = √ (16 + 16)

SR = √32

SR = 4√2

having the values ​​of 2 of its sides we multiply them and obtain their area

PS * RS = Area

2√2 * 4√2 =

16

3 0
3 years ago
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