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djverab [1.8K]
3 years ago
7

please help its says solve 7 3/4 + 2/4 then it says Reduce all fractions to their lowest terms. 12 5/4 12 1/4 11 5/4 11 6/4 11 1

/4
Mathematics
1 answer:
Maru [420]3 years ago
5 0

Answer:

8 1/4 or 33/4

Step-by-step explanation:

7 3/4=31/4

31/4+2/4=33/4

You might be interested in
Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6-ft-high net. It leaves the point of impact wi
Natali [406]

Answer:

a) the vector equation is r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) Maximum height = 7.945ft

c) Time of flight = 1.201 secs and range = 37.45ft

d) when t = 0.74, the distance above the ground = 14.37ft

When t = 0.254, the distance above the ground = 26.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft

Step-by-step explanation:

a) From the diagram

x0 = 0

y0= 4

V0(initial velocity) = 35

α= 27°

y = 4 + 35sin27°t - 16t^2

y = -16t^2 + 35sin27°t+ 4

y = 16t^2 - 35sin27°t - 4

x = 35cos27°t

Vector equation for the path of the volleyball = r(x,y)

r(x,y) = (35cos27°t, 16t^2 -35sin27° - 4)

b) the ball reaches maximum when dy/dt = 0

y = 16t^2 - 35sin27°t - 4

dy/dt = 32 - 35sin27°

0 = 32 - 35sin27°t

-32 = -35sin27°t

t = -32 / -35sin27°

t = 0.4966 seconds

Put t = 0.4966 into y = 16t^2 - 35sin27°t - 4

y = 16(0.4966)^2 - 35sin27°(0.4966) - 4

y = 7.945 ft

The maximum height is 7.945 ft

c) To find the range and flight time, put y= 0

0 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 15.89t - 4

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = -4

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(-4)] /2(16)

= [15.89 +/- √(15.89)^2 + 256] / 32

= (15.59 +/- 22.55) / 32

= (15.89 + 22.55) / 32 or (15.89 - 22.55) / 32

= 1.201 or -0.208

Time of flight = 1.201 secs

Range = x = 35cos27°t

Range = 32cos27°(1.201)

= 37.45 ft

d) when the volley is 7ft above ground, y = 7

Recall that y = 16t^2 - 35sin27°t - 4

7 = 16t^2 - 35sin27°t - 4

0 = 16t^2 - 35sin27°t - 4 +7

0 = 16t^2 - 35sin27°t + 3

0 = 16t^2 - 15.89t + 3

Using quadratic equation general formula,

[-b +/- √b^2 -4ac] / 2a

a = 16, b = -15.89, c = 3

= [-(-15.89) +/- √ (-15.89)^2 - 4(16)(3)] /2(16)

= [15.89 +/- √(15.89)^2 - 192] / 32

= (15.59 +/- 7.7778) / 32

= (15.89 + 7.7778) / 32 or (15.89 - 7.7778) / 32

= 0.74 or 0.254

When t = 0.74,

x = 35cos27°t

x = 35cos27°(0.74)

x = 23.08 ft

Therefore , R - x(0.74)

= 37.45 - 23.08

= 14.37ft

When t = 0.254,

x = 35cos27°t

x = 35cos27°(0.254)

x = 7.92 ft

Therefore R - x(0.254) =

37.45 - 7.92 = 29.53ft

e) if the ball is raised to 8ft, the ball won't be able to reach it. This is because the maximum height is 7.954ft

7 0
3 years ago
4(2x+3) can you help me with this problem
Valentin [98]
Yes I can, MissSmartiez at your service ;)!

This calls for the distributive property. Distributive property, essentially, takes the outside value (in this case 4) and multiply it to the inside value(s) (in this case 2x and [+]3) Thus, it's basically 4 * 2x + 4 *3.

4 multiplied by 2x is 8x. 
4 times 3 is equivalent to 12. 

Therefore the solution is 8x + 12.
3 0
4 years ago
Read 2 more answers
3(4+6)-16 what does it equal and how did you get the answer
Alecsey [184]

Answer:

14

Step-by-step explanation:

you can do this two ways:

<u>First: </u>

3(10)-16\\=30-16\\=14

<u>Second: </u>

Distribute the 3 first

\\12+18-16\\=30

5 0
3 years ago
Maxwell purchased $15,000 worth of 52-week T-Bills for $14,650. What will be the rate of return on his investment?
Ket [755]

Answer:

2.39%

Step-by-step explanation:

Maxwell bought $15,000 worth of 52-week T-bills for $14,650

Therefore the rate of return on investment can be calculated as follows

= 15,000-14,650/14,650× 100

= 350/14,650 × 100

= 0.02389 × 100

= 2.39%

Hence the rate of return on this investment is 2.39%

5 0
3 years ago
A city planner is rerouting traffic in order to work on a stretch of road. The equation of the path of the old route can be desc
Alborosie

Answer:

The correct option is;

y - P = two fifths (x - Q).

Step-by-step explanation:

The equation for the old route is given as follows;

y = \dfrac{2}{5} \cdot x - 4

Therefore, we have;

The slope of the equation of the old route = 2/5

Given that the new route is parallel to the old route and passes through the point (Q, P), the slope of the new route = 2/5 and the equation of the new route in slope and intercept form can be written as follows;

y - P = 2/5×(x - Q)

Therefore, the correct option is y - P = two fifths (x - Q).

6 0
3 years ago
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