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2 years ago

Write the equation of the line that passes through the points (6, 3) and (-7,9). Put

Mathematics

1 answer:

satela [25.4K]2 years ago

4 0

Answer:

y-3= -6/13 ⋅ (x-6)

Step-by-step explanation:

I hope this helps

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(x-1)(x+1)= выполнить умножение

Illusion [34]

Answer:

x^2-1^2

Step-by-step explanation:

formula:

(a-b)(a+b)=a^2-b^2

4 0

1 year ago

Use the distance formula to find the distance to the nearest 10th from R(7,-2) to w(-2,3).

Jlenok [28]

Answer:

10.3

Step-by-step explanation:

d = sqrt ((x2 - x1)^2 + (y2 - y1)^2)

(7,-2)....x1 = 7 and y1 = -2

(-2,3)...x2 = -2 and y2 = 3

now we sub

d = sqrt ((-2 - 7)^2 + (3 - (-2)^2)

d = sqrt ((-9^2) + (5^2))

d = sqrt (81 + 25)

d = sqrt (106)

d = 10.295 rounds to 10.3 <===

7 0

2 years ago

-4-8 and 3-(-1)<br> thats it

mr Goodwill [35]

Answer:

-4-8=-12. 3-(-1)=3

Together they equal -9

Step-by-step explanation:

You have to multiply by the - sign outside the parentheses and the first one just count backwards

8 0

2 years ago

Read 2 more answers

Find the area of the figure

maria [59]

Step-by-step explanation:

Area =0.5×base×height

Area= 0.5×5×3=7.5ft

5 0

2 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago