Solve the equation below 26=8+v v=?

Find the arc length of ADC. Round to the nearest hundredth. BA = 6.5 cm

MakcuM [25]

2190.5 cm.

360 degrees - 23 degrees = 337 degrees

337 degrees x 6.5 cm = 2190.5 cm

7 0

2 years ago

Solve and then use estimation to make sure your answer is reasonable.

olga2289 [7]

Assuming that he read at the same speed last summer, 2,724 pages.

3 0

2 years ago

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

2 years ago

Is 49x^2 a perfect square

Inessa [10]

Yes, this is a perfect square because it can be square rooted to get 7x

3 0

2 years ago

Read 2 more answers

a survey conducted in a school showed that 65% of the students walked to school. The remaining students took buses to school. Th

Feliz [49]

I fear that there is an error copying your assignment.
65% of the students walked. Are all of the other students on buses? Are there both public buses AND private buses?
Assuming that you need to know BOTH kinds of buses, try this:

65% of the students walked, so since 100% - 65% = 35% then this means that 35% of the students were on buses.
Since we know that there are 360 more walkers than bus riders, then one equation we know is: 65% of S = 360 + 35% of S (let S = total # of students)
.65 S = 360 + .35 S
 - .35 S = - .35 S Subtract .35 S from both sides
 .30 S = 360 Divide both sides by .30 (or .3)
.30 .30
S = 1,200 so we know that this is the total number of students, but that is not what was asked.
They want to know how many are on buses and specifically how many are on public buses, if I read this correctly.
Since the walkers = 65% of 1,2000 and we know of means TIMES, then
.65 (1,200) = 780 walkers
1,200 total students minus 780 walkers = 420 bus riders

Now, if there is not a misprint and we really have to figure out the public bus riders as compared to the private bus riders, then remember the ratio from above in the question: 4 bus: 3 public buses
Now if I read this right, that means that 3/4ths of the bus riders were on public buses
so 3/4 of 420 means 3/4 times 420 = 3 times 105 = 315 public bus riders (which coincidentally leaves 105 private bus riders, but since they are private we don't know much about them. Ha-Ha..... I made a lame joke.)

So your answer is 315 public bus riders

5 0

2 years ago

Read 2 more answers

Solve the equation below 26=8+vv=?​

Solve the equation below 26=8+vv=?