Simplify (5 − 3i) + (4 − 8i). (6 points)

If it take Tracy 1/4 of an hour to do her homework,and it takes Trent 3/4 of an hour to do his homework,how much total time does

NeX [460]

It takes them 45 minutes if they start at the same time, but adding them together it would be an hour.

7 0

3 years ago

Read 2 more answers

1. A rabbit can hop 6 feet in 2 seconds and 12 feet in 4 seconds. If the number of feet varies directly with the time the rabbit

kykrilka [37]

Answer: The rabbit is hopping at 3 feet/ sec in speed

Step-by-step explanation:

Step 1

Let the The number of feet of the rabbit be expressed as N

AND time rabbit spends hopping can be expressed as t

From the question, The number of feet varies directly with the time rabbit spends hopping can be expressed as

N α t

Let us introduce a constant of proportionality ,S which represents speed in this case, we now have that

N = St

STEP 2 -- SOLVING

If the rabbit hop 6 feet in 2 seconds, Then speed =

N = St

6= S X2

S = 6/ 2

S = 3

also

If the rabbit hop 12 feet in 4 seconds, Then speed =

N = St

12= S X 4

S = 12/ 4

S = 3

Also recall

that Speed = Distance / time

Pluging the values in the first case,

speed = 6 feet/ 2 secconds = 3 feet/ seconds

in the second case

Speed= 12 feet/ 4 second s= 3 feet/ secs.

5 0

2 years ago

Proportional relationships, please answer all of the parts! Thank you!

Natalka [10]

Answer:

a. No (see below)

b. (see below)

c. Johnny

d. Johnny = $12.50 / hour Doug = $9.16... / hour

e. Doug: extend the table to show how much he gets paid for 1 hour of work (the unit rate)

Johnny: already on the graph

Step-by-step explanation:

a. No. This is because a proportional relationship would have a certain ratio between payment and the amount of time spent working, which would not change. However, if Doug's neighbour pays him $55 for 6 hours, it's likely that this is not a proportional relationship, or we would be able to divide $55 by the 6 hours Doug spent working.

b. Even though above I said that it's probably not a proportional relationship, if it were, we would have to divide everything equally. I'm guessing the table should look like this:

Doug

Time (hr) Earnings ($)

4 $36.66...

5 $45.8333...

6 $55

7 $64.4666....

8 $73.333...

To work out how much Doug would get for working 4 hours:

4/6 x $55 = 2/3 x $55 = $36.66... (so definitely not a proportional relationship)

To work out how much Doug would get for working 5 hours:

5/6 x $55 = $45.8333.... (more evidence that this is not a proportional relationship)

To work out how much Doug would get for working 7 hours (we already got 6 hours):

7/6 x $55 = $64.4666...

To work out how much Doug would get for working 8 hours:

8/6 x $55 = $73.33... (we could double the 4 hour payment, but this is probably more accurate)

c. Johnny is getting paid more. If we look at how much Doug gets for 6 hours, it's $55. If we look out at how much Johnny gets for 6 hours, it's $75. (If we do the same with the other amounts of time, it's the same.

4 hours: D = $36.67 vs J = $50

5 hours: D = $45.83 VS J = $60

etc...

d. Johnny gets $75 per 6 hours --> 75/6 = $12.50 per hour

Doug gets $55 per 6 hours --> 55/6 = $9.1666... per hour

e. Carry on the table for Doug all the way to 1, so you can see the how much Doug gets paid for 1 hour of work (i.e. the minimum).

The unit rate is already on the graph for Johnny.

HOPE THIS HELPED :)

P.S. Mate cld you please give me Brainliest? That table took so much dam work

7 0

3 years ago

Read 2 more answers

You lose 2 points every time you forget to write your name on a test. You have forgotten to write your name 4 times. What intege

Fed [463]

Answer:

The answer is -8

I hope this helps you :D

5 0

2 years ago

Read 2 more answers

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers