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ozzi
4 years ago
8

60,488 round to the nearest ten thousand

Mathematics
2 answers:
svp [43]4 years ago
5 0
60,000 is the correct answer
max2010maxim [7]4 years ago
3 0
60,488 rounded to the nearest 10,000 would be " 60,000 "
You might be interested in
Is it possible to draw a triangle with side lengths 4cm 7cm And 9cm
Lana71 [14]

Answer:

it is

possible

Step-by-step explanation:

<u>sum of the two shortest sides must be greater than the third side.</u>

so

4+7 < 9

11<9

condition satisfies ,

hence proved , we can draw

7 0
3 years ago
Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units
Lady_Fox [76]

Answer:

a. the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b. S=7 m

c. distance=71m

Step-by-step explanation:

a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:

3t^{2}-24t+36=0

so let's solve this for t:

3(t^{2}-8t+12)=0

t^{2}+8t+12=0

and now we factor it again:

(t-6)(t-2)=0

so we get the following answers:

t=6  and t=2

so now we can build our possible intervals:

[0,2)  (2,6)  (6,\infty)

and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:

[0,2) test value t=1

so:

v(1)=3(1)^{2}-24(1)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

(2,6) test value t=3

so:

v(1)=3(3)^{2}-24(3)+36

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

(6,\infty) test value t=7

so:

v(1)=3(7)^{2}-24(7)+36

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

the motion is positive in the time intervals: [0,2)U(6,\infty)

   The motion is negative in the time interval: (2,6)

b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:

s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))

s=7 m

for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:

[0,2)

s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}

so now we evaluate the integral:

s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}

so now we evaluate the integral:

s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}

which simplifies to:

s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}

so now we evaluate the integral:

s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

7 0
3 years ago
F (x + 3)2 = 42, what is the approximate positive value of x?
DaniilM [7]

Answer:

3.48

Step-by-step explanation:

I took the test.

4 0
4 years ago
What are the roots of the equation 9x^2 + 54x +82 = 0 in simplest a + bi form?​
s344n2d4d5 [400]

Answer:

The roots are

<h3>x =  - 3 +  \frac{1}{3}  i \:  \:  \: or \:  \:  \: x =  - 3 -  \frac{1}{3} i \\</h3>

Step-by-step explanation:

9x² + 54x + 82 = 0

Using the quadratic formula

That's

<h3>x =  \frac{ - b\pm \sqrt{ {b}^{2}  - 4ac} }{2a}</h3>

From the question

a = 9 , b = 54 , c = 82

Substitute the values into the above formula and solve

That's

<h3>x =  \frac{ - 54\pm \sqrt{ {54}^{2}  - 4(9)(82)} }{2(9)}  \\ x =  \frac{ - 54\pm \sqrt{2916 - 2952} }{18}  \\ x =  \frac{ - 54\pm \sqrt{ - 36} }{18}  \\ x =  \frac{ - 54\pm 6i}{18}</h3>

Separate the real and imaginary parts

That's

<h3>x =  -  \frac{54}{18} \pm \frac{6}{18}  \: i \\ x =  - 3\pm \frac{1}{3} i</h3>

We have the final answer as

<h3>x =  - 3 +  \frac{1}{3}  i \:  \:  \: or \:  \:  \: x =  - 3 -  \frac{1}{3} i \\</h3>

Hope this helps you

5 0
3 years ago
three neon lights are turnd on at the same time one blinks every 5 second the second every 12 and the third every 6 seconds in 5
mylen [45]

Answer:

5 times

Step-by-step explanation:

If we have to determine the number of times the lights will blink at the same time, we have to find Least common multiple (L.C.M) of 5, 12, and 6 seconds.

5 = 1 x 5

12 = 1 x 2 x 2 x 3

6 = 1 x 2 x 3

The L.C.M of 5, 6, and 12 is = 2 x 2 x 3 x 5 = 60

The three lights will be turned on after 60 seconds or 1 minute.

The lights will be turned on in 1 minute       1 time

The lights will be turned on in 5 minutes    (1*5) times = 5 times

6 0
3 years ago
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