All categories

3 years ago

What percent of a circle is 1/4 + 1/2 ?

Mathematics

2 answers:

zlopas [31]3 years ago

7 0

Presuming the circle is the whole, 1/4 + 1/2 is 3/4, or 0.75, so it is 75% of the circle.

vesna_86 [32]3 years ago

4 0

Well 1/4 is .25 and 1/5 is .5 so it would be .75 so 75%? I'm not sure if that's right, but I hope I helped!

You might be interested in

Does 3 work as a solution to the following inequality? 2n > 6

Ksju [112]

Answer:

No.

Step-by-step explanation:

If n=3 the only thing you get is 6, which is not more than 6.

5 0

3 years ago

Which number is closest to 66 (squared)? A) 8.8 B) 8.1 C 8.2 D) 9.1

Hoochie [10]

Answer:

sq rt of 66 is 8.12 so 8.1 closest

8 0

3 years ago

Read 2 more answers

PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwi

il63 [147K]

To solve our questions, we are going to use the kinematic equation for distance: $d=vt$
where
$d$ is distance
$v$ is speed
$t$ is time

1. Let $v_{w}$ be the speed of the wind, $t_{w}$ be time of the westward trip, and $t_{e}$ the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/hr, so $v=450$ . Now we can use our equation the relate the unknown quantities with the quantities that we know:

Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
 $d=vt$
$2400=(450-v_{w})t_{w}$

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
$d=vt$
$2400=(450+v_{w})t_{e}$

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate: $450-v_w$ Time: $t_w$
Going eastward -Distance: 2400 Rate: $450+v_w$ Time: $t_e$

2. Notice that we already have to equations:
Going westward: $2400=(450-v_{w})t_{w}$ equation(1)
Going eastward: $2400=(450+v_{w})t_{e}$ equation (2)

Let $t_{t}$ be the time of the round trip. We know from our problem that the round trip takes 11 hours, so $t_{t}=11$ , but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so $t_{t}=t_w+t_e$ . Using this equation we can express $t_w$ in terms of $t_e$ :
$t_{t}=t_w+t_e$
$11=t_w+t_e$ equation
$t_w=11-t_e$ equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns:
$2400=(450-v_{w})t_{w}$
$2400=(450-v_{w})(11-t_e)$

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

3. Lets solve our system of equations to find the speed of the wind:
$2400=(450-v_{w})(11-t_e)$ equation (1)
$2400=(450+v_{w})t_{e}$ equation (2)

Step 1. Solve for $t_{e}$ in equation (2)
$2400=(450+v_{w})t_{e}$
$t_{e}= \frac{2400}{450+v_{w}}$ equation (3)

Step 2. Replace equation (3) in equation (1) and solve for $v_w$ :
$2400=(450-v_{w})(11-t_e)$
$2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )$
$2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} )$
$2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}}$
$2400(450+v_{w})=1147500+2400v_w-11(v_w)^2$
$1080000+2400v_w=1147500+2400v_w-11(v_w)^2$
$(11v_w)^2-67500=0$
$11(v_w)^2=67500$
$(v_w)^2= \frac{67500}{11}$
$v_w= \sqrt{\frac{67500}{11}}$
$v_w=78$

We can conclude that the speed of the wind is 78 mi/hr.

6 0

3 years ago

Weights of babies at the local community hospital have a distribution that is approximately normal with a mean weight of 7.43 po

KengaRu [80]

Answer:

Step-by-step explanation:

Hope this helped