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2 years ago

1. The point A(-4,6) is rotated 90 degrees counterclockwise and then translated 3 units up.

Mathematics

1 answer:

professor190 [17]2 years ago

5 0

Answer:

Step-by-step explanation:

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Rida and elisa are each paid by the hour. rida worked 5 1\2. hours and earned 77. elisa worked 3 3\4 HOURS AND EARNED 60. WHOSE

skad [1K]

Rida:
5 1/2 :77
5.5 : 77
Divide both sides by 5.5 to get $/hour:
1:14

Elisa:
3 3/4:60
3.75:60
Divide both sides by 3.75 to get $/hour:
1:16

16 is bigger than 14 so Elisa earns more.

Hope this helps :)

5 0

3 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago

Identify the segments that are parallel, if any, if ∠ADH≅∠ECK.

Dafna1 [17]

9514 1404 393

Answer:

C. none of these

Step-by-step explanation:

The given information tells us ΔACD is isosceles, but gives no information about any lines that might conceivably be parallel.

3 0

2 years ago

How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you

kupik [55]

I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6

you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5

II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7

then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15

3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1

insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4

insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2

so the solution is
x=2
y=-4
z=-1

3 0

3 years ago

There are 8 roses in a vase of 13 flowers. The rest are daisies.

Blizzard [7]

Answer:

5:13
8:5

Step-by-step explanation:

For first part:

Roses = 8
Daises = All flowers - Roses
Daises = 13 - 8
Daises = 5
Daises:All Flowers = 5:13

For second part:

Roses:Daises = 8:5

I hope this helps!

7 0

2 years ago

Read 2 more answers