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Setler [38]
3 years ago
15

Carl delivers papers for spending money. He gets $10

Mathematics
1 answer:
postnew [5]3 years ago
5 0

Answer:30 papers

Step-by-step explanation:

if he works one day that means he makes $10 , which leaves $15 , divide $15 by $0.50.

that gives you 30 papers

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6x-1 + 4X+21 = 50 solve for x
emmainna [20.7K]
Answer:

x=3


Explanation:

First what you do is move x to one side and the rest of the numbers to another while changing the sign from + to - or - to + like this
6x+4x= 1+-21+50

Then you do the equation on 1 side
6x+4x=30

And then the other but add them like x is not there
10x=30
Your final result should be
X=3
3 0
3 years ago
Read 2 more answers
Two point particles are attached to each other via a light in-extensible string that passes over a ring hanging from the ceiling
arlik [135]

a. The expression for the tension in the string is T = mg

b. The expression for the tension in the chain is T = 2mg

To solve the problem, we need to know what tension is.

<h3>What is tension?</h3>

This is the opposing force in a stretched material.

<h3>a. Tension in the string</h3>

The expression for the tension in the string is T = mg

Since the string supports the weight of the object, the tension in the string equals the weight of the particle.

Let

  • T = tension in sring and
  • W = weight of particle = mg where
  • m = mass of particle and
  • g = acceleration due to gravity

So, T = W

T = mg

So, the expression for the tension in the string is T = mg

<h3>b. Tension in chain</h3>

The expression for the tension in the chain is T = 2mg

<h3 />

Since the chain supports both strings, the tension in the chain equals the tension in both strings.

Let

  • T' = tension in chain and
  • T = tension in each string

So, T' = T + T

T' = 2T

T' = 2W

T' = 2mg

So, the expression for the tension in the chain is T = 2mg

<h3 />

Learn more about tension here:

brainly.com/question/24994188

5 0
2 years ago
Directions: Decide which measurement is larger, by converting to the same measurement.
atroni [7]

Answer:

9 yards.

Step-by-step explanation:

3 feet=1 yard.

9x3=27.

27>25.

So in conclusion, 9 yards is the bigger measurement.

8 0
2 years ago
Read 2 more answers
During halftime of a basketball ​game, a sling shot launches​ T-shirts at the crowd. A​ T-shirt is launched from a height of 4 f
Sonbull [250]

Answer:

(a) The time the T-shirt takes to maximum height is 2 seconds

(b) The maximum height is 68 ft

(c) The range of the function that models the height of the T-shirt over time given above is 4 + 64\cdot t - 16 \cdot  t^{2}

Step-by-step explanation:

Here, we note that the general equation representing the height of the T-shirt as a function of time is

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

Where:

h = Height reached by T-shirt

t = Time of flight

u = Initial velocity = 64 ft/s

g = Acceleration due to gravity (negative because upward against gravity) = 32 ft/s²

h₁ = Initial height of T-shirt = 4 ft

(a) The maximum height can be found from the time to maximum height given as

v = u - gt

Where:

u = Initial velocity = 64 ft/s

v = Final upward velocity at maximum height = 0 m/s

g = 32 ft/s²

Therefore,

0 = 64 - 32·t

32·t = 64 and

t = 64/32 = 2 seconds

(b) Therefore, maximum height is then

h = 4 + 64\times 2 - \frac{1}{2} \times 32  \times  2^{2}

∴ h = 68 ft

The T-shirt is then caught 41 ft above the court on its way down

(c) The range of the function that models the height of the T-shirt over time given above is derived as

h = h_1 + u\cdot t - \frac{1}{2} \cdot g  \cdot  t^{2}

With u = 64 ft/s

g = 32 ft/s² and

h₁ = 4 ft

The equation becomes

h =4 + 64\cdot t - \frac{1}{2} \times 32  \cdot  t^{2} = 4 + 64\cdot t - 16 \cdot  t^{2}.

6 0
3 years ago
If the diagonals of a quadrilateral bisect the angles, is the quadrilateral always a parallelogram? Explain your answer.
Liula [17]
Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.


Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).

Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.
6 0
3 years ago
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