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never [62]
3 years ago
12

. Write three numbers that round to 38,000 when rounded to the nearest thousand.

Mathematics
2 answers:
Vlad [161]3 years ago
8 0

Answer:

i. 37, 788

ii. 38, 222

iii. 37, 900

Step-by-step explanation:

Rounding number  can be very easy to work with in the head. The rounded number is the approximate value.To round a number to the nearest thousand simply means making the last three digit into the next lower number that ends in 000. Example round 6897 to the nearest thousand . The last three digits 897 depict that 6897 is more closer to 7000 than 6000 so, the nearest thousand of that value (6897) will be 7000.

The number can be either rounded up or down to the closest thousand depending on the closest to the nearest thousand. 6347 tot the closest thousand will be 6000 . Notice that the last three digits(347) indicate that the number (6347) is more closer to 6000 than 7000.

From the example above the three numbers that can be rounded to 38 000 are

i.  37, 788

ii. 38, 222

iii. 37, 900

stiks02 [169]3 years ago
5 0

Answer:

37999, 38100, 38300

Step-by-step explanation:

37999

38100

38300

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3 years ago
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 73 and 9
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Step-by-step explanation:

The sinusoidal function is given by :

y=A\sin[\omega(x-\alpha)]+C

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As per given,

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Phase shift: \alpha=10

Period = 24 hours;

\omega=\dfrac{2\pi}{24}=\dfrac{\pi}{12}

Substitute all values in sinusoidal function, we get

y=12\sin[\dfrac{\pi}{12}(x-10)]+85

Put y= 82, we get

82=12\sin[\dfrac{\pi}{12}(x-10)]+85\\\\\Rightarrow\ -3= 12\sin[\dfrac{\pi}{12}(x-10)]\\\\=\dfrac{-1}{4}= \sin[\dfrac{\pi}{12}(x-10)]\\\\\Rightarrow\ \dfrac{\pi}{12}(x-10)=\sin^{-1}(\dfrac{-1}{4})\\\\\Rightarrow\ x-10=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))\\\\\Rightarrow\ x=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))+10\\\Rightarrow\ x\approx9.03

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3 years ago
Jessie draws triangle ABC on a coordinate grid. The slope of line segment AB is Jessie then transforms triangle
balu736 [363]

Answer:

1) Supports Jessie's Claim

2) Does Not Support Jessi's Claim

3) Supports Jessie's Claim

4) Does Not Support Jessi's Claim

Step-by-step explanation:

The given transformations are;

1) Rotation of 180° around the origin

For a rotation of 180° around the origin, either clockwise or anti clockwise, for a given coordinate of the preimage (x, y), the coordinate of the image is (-x, -y)

Therefore, whereby the slope of the preimage, given two points (0, 0) and (2, 2), = (2 - 0)/(2 - 0) = 1

For the image with the points (0, 0) and (-2, -2), we have;

(-2 - 0)/(-2 - 0) = 1

Therefore, the slope of the preimage and the image are equal

Therefore, supports Jessie's Claim

2) For a reflection across the line y = 2, we have

We note that the line y = 2 is parallel to the x-axis

For a reflection across the x-axis, for a preimage (x, y), we have the coordinates of the image (x, -y)

However for the reflection across the line y = 2, we have;

For a preimage, (x, y), the coordinate of the image is (x, -y+4)

Given two points, of the preimage (0, 0) and (2, 2), we have the image given as (0, 4) and (2, -2 + 4) = (2, 2);

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (2 - 4)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

3) For a translation up 1.25 units, we note that the difference in the y and x values of the coordinates of the preimage and the image will be equal when finding the slope, and therefore, the slope of the figure of the preimage and the slope of the figure of the image will be equal

Therefore, supports Jessie's Claim

4) For a reflection across the x-axis, a point on the preimage, with coordinates (x, y) will form a point on the image with coordinates (x, - y)

For a preimage with points (0, 0) and (2, 2), we have the image as (0, 0) and (2, -2)

The slope of the preimage is (2 - 0)/(2 - 0) = 1

The slope of the image is (-2 - 0)/(2 - 0) = -1

The slope of the line of the preimage and the image are different

Therefore, does Not Support Jessi's Claim

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3 years ago
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