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Aliun [14]
3 years ago
11

What is the sum of 2/3 and 3/4

Mathematics
2 answers:
Mumz [18]3 years ago
8 0
2/3 + 3/4 = 17/12 or 1 5/12
oksano4ka [1.4K]3 years ago
3 0
The sum is 2/3 +3/4=8/12 +9/12 =17/12
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Im so confused PLEASE HELP ASAP
siniylev [52]

Answer:

adding 2 to each number

Step-by-step explanation:

7 +2 = 9 + 2 = 11

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The graph of a linear equation passes through three of these points: (0, 5), (2, 2), (3, 1), and (4, -1). The point that the gra
IgorLugansk [536]

Answer:

(3,1) is the answer your looking for.

Step-by-step explanation:

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ANSWER THE QUESTION IN THE PIC. WILL GIVE BRAINLIEST!
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Slope and y intercept
morpeh [17]

Answer:

y=\frac{25}{1}x+75

The beginning balance of Sophie's account is 75

Step-by-step explanation:

Using the slope intercept form aka y=mx+b

The slope of a line describes how steep a line is.

  • We can see that the y-axis goes up by 25.
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<u>$ is how much you have in your bank account.</u>

<em>We can see that Sophie starts off with 75 dollars.</em>

In the first week, we see that Sophie has 100 dollars. In the second week, we see that Sophie has 125.

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slope formula\\m=\frac{rise(y)}{run(x)}=\frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

The first week (1, 100)

The second week (2,125)

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<em></em>

7 0
3 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
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