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Rainbow [258]
3 years ago
8

Round 1,287 to the nearest thousands place

Mathematics
2 answers:
Dmitry [639]3 years ago
5 0

Answer:

1,000

Step-by-step explanation:

refer to attached.

the number in the thousands place is the number 1.

How we round this number depends on the number immediately to the right of it (i.e the hundreds number)

Case 1: If the number in the hundreds place is less than 5, then we keep the number in the thousands place the same and replace all the digits to the right of the thousands digit by zeros

Case 2: If the number in the hundreds place is 5 or more, then increase the number in the thousands place  by 1 and replace all the digits to the right of the thousands digit by zeros

in our case, the hundreds digit is 2, which falls into case 1 above,

hence according to the procedure in case 1 above, we keep number 2 the same and replace everything to the right of 2 with zeros, giving us

1,000

alexdok [17]3 years ago
3 0

Answer:

1000

Step-by-step explanation:

If a number is higher than 5 round up. If its 4 or lower round down

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Expanded form is writing out everything and no simplifying.

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Solve for x in the equation;3x +34x -40x=0
saveliy_v [14]

Answer:

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Step-by-step explanation:

3x + 34x - 40x = 0

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3 years ago
Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and
zzz [600]

Answer:

\bold{sin(38^\circ)=cos(52^\circ)}

Step-by-step explanation:

Given that \triangle KJL is a right angled triangle.

\angle JKL = 52^\circ\\\angle KLJ = 38^\circ

and

\angle KJL = 90^\circ

Kindly refer to the attached image of \triangle KJL in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given \triangle KJL.

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

sin\theta =\dfrac{Perpendicular}{Hypotenuse}

sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

cos\theta =\dfrac{Base}{Hypotenuse}

cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}....... (2)

By (1) and (2):

sin(38°) \neq cos(38°).

cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL} ...... (3)

Comparing equations (1) and (3):

we get the both are same.

\therefore \bold{sin(38^\circ)=cos(52^\circ)}

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Are all vertical lines parallel?
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