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Viktor [21]
3 years ago
14

Please look at number 13. and use the slope and write the equation for me

Mathematics
1 answer:
natta225 [31]3 years ago
7 0

Answer:

y - y1 = m(x - x1)

(x1, y1) = (1, -1)  (x2, y2) = (3, 5)

m = \frac{5 - (-1)}{3 - 1}

m = \frac{5 + 1}{3-1}

m = \frac{6}{2} = \frac{3}{1} = 3

m = 3

y - 5 = 3(x - 3)

y - 5 = 3x - 9

y = 3x - 9 + 5

y = 3x - 4

Step-by-step explanation:

In order to use point-slope form to convert to slope-intercept form you have to know one point and the value of the slope. In this case, we know two points, but we don't know the slope. In order to find the slope, you can input the values of the two points into the expression: \frac{y2 - y1}{x2 - x1}

After finding the slope, input the values of one of the points (in this example I used (3,5)) and the slope. After that, multiply the slope by the two terms inside the parentheses. then add 5 to both sides of the equation to isolate y. Now your equation is in slope-intercept form.

I hope this helps :)

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marin [14]
Neither. There are 3/6 chances of getting an even number on a number cube, and 1/2 chances of landing on heads.

3/6 is equal to 1/2, so the chances are the same.



Please consider marking this answer as Brainliest to help me advance.
5 0
4 years ago
Read 2 more answers
Please help, and i will mark you BRAINLIEST <br> Solve: <br> A. 27<br> B. 343<br> C. 729<br> D. 216
TEA [102]

Answer:

729

Step-by-step explanation:

9 x 9 = 81

81 x 9 = 729

8 0
3 years ago
Can someone please help with these 2 word problems ASAP.
PolarNik [594]

Answer:

4.     4.66hours

5.  time_trip = 2.777 h

Step-by-step explanation:

4. A jet plane travels 2 times the speed of a commercial airplane. The distance between Vancouver and Regina is 1730 km. If the flight from Vancouver to Regina on a commercial airplane takes 140 minutes longer than a jet plane, what is the time of a commercial plane ride of this route?

We know that

Speed_commercial = distance/ time_comm

Speed_commercial = 1730 km/ time_comm

Then,

Speed_jet = distance/ time_jet = 2*Speed_commercial

Speed_jet =   1730 km/ time_jet = 2*Speed_commercial

time_comm = 2.333 h + time_jet

1730 km/ time_jet = 2*1730 km/ time_comm

time_comm = 2*time_jet

time_comm = 2*(time_comm  - 2.333h)

time_comm = (2*time_comm  - 4.666h)

time_comm  = 4.666 h

5. A man goes fishing in a river and wants to know how long it will take him to get 10km upstream to his favourite fishing location. The speed of the current is 3 km/hr and it takes his boat twice as long to go 3km upstream as is does to go 4km downstream. How long will it take his boat to get to his fishing spot?

Distance = 10 km upstream

Speed_current = 3 km/h

Upstream

(Speed_boat - Speed_current)  = 3 km / (2*time_downstream)

Downstream

(Speed_boat + Speed_current)  = 4 km / (time_downstream)

We subtract both equations

(Speed_boat + Speed_current) - (Speed_boat - Speed_current) = 4 km / (time_downstream) - 3 km / (2*time_downstream)

2*Speed_current = (5/2 km)/ time_downstream

2*(3 km/h)= (5/2 km)/ time_downstream

time_downstream = 0.416 h

(Speed_boat + 3km/h)  = 4 km / (0.416 h)

Speed_boat  = 6.6 km/h

Trip upstream

(Speed_boat - Speed_current)  = 10 km / (time_trip)

time_trip = 10 km/(3.6 km/h) = 2.777 h

time_trip = 2.777 h

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The two objects are similar because of their gcf

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Answer:

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