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2 years ago

Use the graph to describe what transformation has taken place. List your answer below.

Mathematics

1 answer:

Mrrafil [7]2 years ago

7 0

Answer:

Reflection over y-axis (-x,y)

Step-by-step explanation:

Reflection over y-axis (-x,y)

A (-2 , 4) , A'( 2,4)

you can check the rest.

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Find that image of P(-2,4) under a translation along the vector (6,5)

ollegr [7]

Answer:

P'(4, 9)

Step-by-step explanation:

Add the translation vector to the point:

P' = P +(6, 5) = (-2, 4) +(6, 5) = (-2+6, 4+5)

P' = (4, 9)

4 0

2 years ago

Please help!? Evaluate the expression 5V - 6w when v= 4 and w= 2.

Rudik [331]

Answer:

Step-by-step explanation:

5V-6W

V=4. W=2.

5(4)-6(2)

20-6(2)

20-12=

8 0

3 years ago

A $105 watch is selling for 30% off.Find the final price if the sales tax is 8%

Nesterboy [21]

105 x 0.08 and what you get is your answer

6 0

2 years ago

URGENT HELP ME PLEASE

Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

5 0

3 years ago

at a point P on the parabola x^2=4ay a normal PK is drawn. From vertex O, a perpendicular OM is drawn to meet the normal at M. S

saveliy_v [14]

Answer: Its too long to write here, so I will just state what I did. I let P=(2ap,ap^2) and Q=(2aq,aq^2) But x-coordinates of P and Q differ by (2a) So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2) So Q=(2a(p-1), aq^2) which means, 2aq = 2a(p-1) therefore, q=p-1 then I subbed that value of q in aq^2 so Q=(2a(p-1), a(p-1)^2) and P=(2ap,ap^2) Using these two values, I found the midpoint which was: M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 ) then x = a(2p-1) rearranging to make p the subject p= (x+a)/2a

6 0

2 years ago