29. 83 – 2 · 4 = 30. 3 · 6 – 4 · 2 = 31. 3( 6

b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.

Ainat [17]

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

Step-by-step explanation:

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ $2x$

The third box contained two more pounds than the first box i.e.

⇒ $x+2$

The last box contained one-fourth the amount in the second box i.e.

⇒ $(\frac{1}{4})2x = \frac{x}{4}$

Therefore , Sum of pounds of candy are :

⇒ $\frac{x}{2} +x+2+2x+x=65$

⇒ $\frac{x}{2} +4x=63$

⇒ $\frac{9x}{2}=63$

⇒ $x=63(\frac{2}{9} )$

⇒ $x=14$

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

8 0

3 years ago

Terri played 70 tournament chess games during the school year and won 49 of them. What is her win-to-loss ratio?

Scilla [17]

Alright so the answer would be letter B, because

win-to-loss

49 to 21 (You can determine her losses by subtracted the number of wins from her total number of games played)

Now you must simplify your answer

7 is the greatest common factor (GCF) of both 49 and 21

Now divide both 49 and 21 by 7 to get

7 to 3

7 0

2 years ago

A tour director collected $258 for 13 tour packages tour packages a cost $18 into a package requires $22 how many of each tour p

Lena [83]

Answer:

Step-by-step explanation:

X+Y = 13

$18 + $22 = $258

OR

18X+22Y=258

18X + 22(13 - X) = 258

18X + 286 - 22X = 258

286 - 4X = 258

-4X= 258 - 286

-4X = -28

X= 7

Y=6

7 of the X tour packages & 6 of the Y tour packages

4 0

2 years ago

A man hits a baseball when it is 4 ft above the ground with an initial velocity of 120 ft/sec. The ball leaves the bat at a 30 d

Sauron [17]

et us assume that the origin is the floor right below the 30 ft. fence

To work this one out, we'll start with acceleration and integrate our way up to position.

At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.

To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60

To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4

If the ball clears the fence, it means x = 0, y > 30

x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds

for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft

which is less than 30ft, so it means that the ball will NOT clear the fence.

Just for fun, let's check what the speed should have been :)

x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4

x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s

v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s

So he needed to hit the ball at at least 123.34 ft/s to clear the fence.

You're welcome, Thanks please :)

6 0

2 years ago

Oh for the love of noodles, please help me T-T

nirvana33 [79]

I would say C hope I’m not to late :’)

7 0

2 years ago

29. 83 – 2 · 4 = 30. 3 · 6 – 4 · 2 = 31. 3( 6 – 4) · 2 =