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alex41 [277]
3 years ago
13

SUPER EASY AND BRAINLIEST!!!!!

Mathematics
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

x+3y =4

Step-by-step explanation:

The standard form for a line is Ax +By =C

We have a point and a slope.  Use point slope form to get the line and then transform it to standard form

Point slope form

y-y1 = m(x-x1) where (x1.y1) is the point and m is the slope

Substituting in the point and the slope

y-0 = -1/3(x-4)

y = -1/3 (x-4)

Multiply each side by 3 to get rid of the fraction

3y = 3*(-1/3 (x-4))

3y = -(x-4)

Distribute the minus sign

3y = -x +4

Add x to each side

x+3y = x-x+4

x+3y = 4

This is in standard form

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Urban gardening is becoming popular around the world. Some city residents grow plants in containers, like buckets or pots. Other
ElenaW [278]

Let x = the total number of garden beds

18 = .3x

x = 60 garden beds

7 0
4 years ago
Not sure if any of this is correct, but it’s what I got so far
Irina18 [472]

Problem 1 is correct. You use the pythagorean theorem to find the hypotenuse.

==================================================

Problem 2 has the correct answer, but one part of the steps is a bit strange. I agree with the 132 ft/sec portion; however, I'm not sure why you wrote \frac{1 \text{ sec}}{132 \text{ ft}}=\frac{0.59\overline{09}}{78 \text{ ft}}*127 \text{ ft}

I would write it as \frac{1\text{ sec}}{132 \text{ ft}}*127 \text{ ft} = \frac{127}{132} \text{ sec} \approx 0.96 \text{ sec}

==================================================

For problem 3, we first need to convert the runner's speed from mph to feet per second.

17.5 \text{ mph} = \frac{17.5 \text{ mi}}{1 \text{ hr}}*\frac{1 \text{ hr}}{60 \text{ min}}*\frac{1 \text{ min}}{60 \text{ sec}}*\frac{5280 \text{ ft}}{1 \text{ mi}} \approx 25.667 \text{ ft per sec}

Since the runner needs to travel 90-12 = 78 ft, this means\text{time} = \frac{\text{distance}}{\text{speed}} \approx \frac{78 \text{ ft}}{25.667 \text{ ft per sec}} \approx 3.039 \text{ sec}

So the runner needs about 3.039 seconds. In problem 2, you calculated that it takes about 0.96 seconds for the ball to go from home to second base. The runner will not beat the throw. The ball gets where it needs to go well before the runner arrives there too.

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

The question is now: how much of a lead does the runner need in order to beat the throw?

Well the runner needs to get to second base in under 0.96 seconds.

Let's calculate the distance based on that, and based on the speed we calculated earlier above.

\text{distance} = \text{rate}*\text{time} \approx (25.667 \text{ ft per sec})*(0.96 \text{ sec}) \approx 24.64032 \text{ ft}

This is the distance the runner can travel if the runner only has 0.96 seconds. So the lead needed is 90-24.64032 = 65.35968 feet

This is probably not reasonable considering it's well over halfway (because 65.35968/90 = 0.726 = 72.6%). If the runner is leading over halfway, then the runner is probably already in the running motion and not being stationary.

As you can see, the runner is very unlikely to steal second base. Though of course such events do happen in real life. What may explain this is the reaction time of the catcher may add on just enough time for the runner to steal second base. For this problem however, we aren't considering the reaction time. Also, not all catchers can throw the ball at 90 mph which is quite fast. According to quick research, the MLB says the average catcher speed is about 81.8 mph. This slower throwing speed may account for why stealing second base isn't literally impossible, although it's still fairly difficult.

5 0
3 years ago
Expand each logarithm <br> <br>Log (6*11)
MA_775_DIABLO [31]
Log(66)
log(6 × 11)
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5 0
3 years ago
Use this graph of velocity vs. time for two objects to answer the question.
Lynna [10]

Using derivatives, it is found that the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

The acceleration is the <u>derivative of the velocity</u>, given by change in velocity divided by change in time, that is:

a = \frac{\Delta_v}{\Delta_t}

In this problem:

  • The change in time for objects C and T is the same.
  • The <u>change in velocity for object C is greater</u>, thus, it has a greater acceleration, and the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

A similar problem is given at brainly.com/question/14516604

6 0
3 years ago
Read 2 more answers
Find the range of possible lengths of the third side using the given side lengths <br> 10, 7
Tomtit [17]

Answer:

3 < x < 17

Step-by-step explanation:

Given 2 sides then the possible range of the third side x is

difference of 2 sides < x < sum of 2 sides , that is

10 - 7 < x < 10 + 7

3 < x < 17

8 0
3 years ago
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