1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nadya68 [22]
3 years ago
14

5.986 to nearest thousands

Mathematics
2 answers:
katovenus [111]3 years ago
7 0

5.986 is already to the nearest thousandth. (see the picture)

ludmilkaskok [199]3 years ago
6 0

5,986 to the nearest thousand is 6,000

To work this out, you look at the hundreds digit (which is 9). If it is between 1 and 4, you round it down, and if it is between 5 and 9, you round it up.

Hope this helps!

You might be interested in
You put 15 gallons of gasoline in your car, you know that this amount of gasoline will allow you to drive about 450 miles?
solmaris [256]

i dont get wht u are trying to say?

3 0
4 years ago
(I'LL GIVE BRAINLIEST TO WHO EVER HELPS ME!!)
ra1l [238]

Answer:

C) 93 r 3

Your Welcome! ^O^

   0 0 9 3

7 2 6 6 9 9

 − 0      

   6 6    

 −   0    

   6 6 9  

 − 6 4 8  

     2 1 9

   − 2 1 6

         3

7 0
3 years ago
A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$$

$$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$$

$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}

$$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$$

$$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$$

$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square

8 0
2 years ago
25 = 150 * (0.997)^t<br><br> solve without using logs
V125BC [204]

Answer:

hi hi hi hi hi

Step-by-step ms. Hi hi hi hi

5 0
3 years ago
A ray of light with intensity lo falls vertically on the sur face of the sea and at ten meters deep its intensity is 10/3. Assum
seropon [69]

Answer:

intensity = \frac{Io}{15}

intensity = \frac{Io}{30}

depth = 333.33 m

Step-by-step explanation:

given data

deep = 10 m

intensity = Io

intensity = Io/3

to find out

intensity of light at 50 m and 100 m and 1/100 of initial intensity remain ?

solution

we know here that intensity is inversely proportional to deep so

intensity = k × \frac{1}{Deep}      .................1

here k is constant

so we have given 10 m deep so

\frac{Io}{3}  = \frac{k}{10}

so k = Io × \frac{10}{3}    ................2

so from equation 1 when 100 m deep and 50 m deep

intensity = k × \frac{1}{Deep}

intensity =  Io × \frac{10}{3} × \frac{1}{50}

intensity = \frac{Io}{15}

and

intensity =  Io × \frac{10}{3} × \frac{1}{100}

intensity = \frac{Io}{30}

and

at intensity Io/100

intensity = k × \frac{1}{Deep}

\frac{Io}{100}  =  Io × \frac{10}{3} × \frac{1}{D}

D = 333.33 m

so depth = 333.33 m

3 0
3 years ago
Other questions:
  • What number can you add to π to get a rational number?
    11·2 answers
  • Simplify (23 × 43)². Express as a single power.
    7·1 answer
  • What is the difference between a line graph and a scatter plot? a. A line graph presents continuous and linked data, while a sca
    9·2 answers
  • What is the expression of 2+32times5
    5·2 answers
  • The wavelength of violet light is 4.0 x 10−7 m. The wavelength of red light is 6.5 x 10−7 m. How much longer is the wavelength o
    13·1 answer
  • The function f is such that f(x) = x^2 – 2x + 3
    14·2 answers
  • A circular garden has a diameter of 8 yards. What is the area of the garden? Use 3.14 for π.
    12·1 answer
  • Which statement describes the line that contains the points (5, -6) and (3, -6)?
    14·1 answer
  • Please answer correctly math problemo
    6·2 answers
  • PLEASE HELP FAST!!!!
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!