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Rainbow [258]
3 years ago
7

What’s the equation of a line that passes through (-6,5) (-3,-3)

Mathematics
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

<h3>y=-8/3x-11</h3>

Step-by-step explanation:

<em><u>SLOPE FORMULA:</u></em>

\Rightarrow \displaystyle \mathsf{\frac{Y_2-Y_1}{X_2-X_1} }}

Y2=(-3)

Y1=5

X2=(-3)

X1=(-6)

Solve.

\displaystyle \mathsf{\frac{-3-5}{-3-\left(-6\right)}=\frac{8}{3}=-\frac{8}{3}  }}}

Then, compute by the y-intercept.

<u><em>SLOPE INTERCEPT FORM FORMULA:</em></u>

\displaystyle \mathsf{Y=MX+B}}

y=(-8/3)x+b

(-6,5)

X=(-6)

Y=5

5=(-8/3)(-6)+b

Isolate b on one side of the equation.

(-8/3)(-6)+b=5 (Switch sides.)

Remove parenthesis.

8/3*6+b=5

8*6/3

Multiply.

8*6=48

Divide.

48/3=16

16+b=5

b+16=5 (Switch sides.)

Then, subtract 16 from both sides.

b+16-16=5-16

Solve.

5-16=-11

b=-11

So, the correct answer is y=-8/3x-11.

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If I have 4 groups of $29, about how much do I have?
grandymaker [24]

Answer:

i think its $116

Step-by-step explanation:

if you have four groups of twenty-nine dollars that would be

4x$29

which is $116

6 0
3 years ago
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You design a tree house using a coordinate plane in which the coordinates are measured in feet. The vertices of the floor are (-
Nikitich [7]

Answer:

Perimeter = 28 yards

Area = 49 square yards.

Step-by-step explanation:

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let as consider the vertices of the floor are A(-2,-3), B(-2,4), C(5,4) and D(5,-3).

Using distance formula, we get

AB=\sqrt{(-2-(-2))^2+(4-(-3))^2}=\sqrt{(-2+2)^2+(4+3)^2}=\sqrt{0+7}^2=7

Similarly,

BC=\sqrt{(5-(-2))^2+(4-4)^2}=7

CD=\sqrt{(5-5)^2+(4-(-3))^2}=7

AD=\sqrt{(5-(-2))^2+(-3-(-3))^2}=7

It is conclude that, AB=BC=CD=AD.

From the given points it is clear that all sides lie on either vertical or horizontal lines.

Since all sides are equal, and adjacent sides are perpendicular, therefore, the base is a square with edge 7 yards.

Perimeter of square floor is

P=4a=4(7)=28\text{ yards}

Area of square floor is

A=a^2=7^2=49\text{ sq. yards}

Therefore, the perimeter is 28 yards and the area is 49 square yards.

7 0
3 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
Could use some help with this. Ty to all who help!
Gekata [30.6K]
106 they add up to 180
6 0
3 years ago
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I don't know if I'm right or not. Can someone please tell me if I'm right please?
Lorico [155]

Answer:

Yes you are right

Step-by-step explanation:

1st of all you said ''tell me if im right please'' =3

7 0
3 years ago
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