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3 years ago

Round 937,003 to the nearest ten thousand

Mathematics

2 answers:

prisoha [69]3 years ago

4 0

Answer:

937000

Step by Step Ex:

zubka84 [21]3 years ago

3 0

Answer:940000

Step-by-step explanation: if number that is right of the number you are rounding is 0-4 the number you are rounding stays the same if the number left of the number you are rounding is a 5-9 then the number you are rounding increases by one and all the numbers right of the number you are round turn to zero. Hope this helps

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Free brainliest whoever can guess my character name??

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Sue made a batch of snack mix for the party. She combined 12 ounces of mixed nuts, 20 ounces of cereal, and 8 ounces of pretzels

Levart [38]

Answer:

The pretzels are 20% of the total weight of the snack.

Step-by-step explanation:

Total weight 12 + 20 + 8 = 40

So 8 / 40 = 20%

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3 years ago

3x+4=7-2x <br> Solve this equation for x with steps

ollegr [7]

Simplifying
3x + 4 = 7 + -2x

Reorder the terms:
4 + 3x = 7 + -2x

Solving
4 + 3x = 7 + -2x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '2x' to each side of the equation.
4 + 3x + 2x = 7 + -2x + 2x

Combine like terms: 3x + 2x = 5x
4 + 5x = 7 + -2x + 2x

Combine like terms: -2x + 2x = 0
4 + 5x = 7 + 0
4 + 5x = 7

Add '-4' to each side of the equation.
4 + -4 + 5x = 7 + -4

Combine like terms: 4 + -4 = 0
0 + 5x = 7 + -4
5x = 7 + -4

Combine like terms: 7 + -4 = 3
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Divide each side by '5'.
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Answer: x = 0.6

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2 years ago

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Sunny_sXe [5.5K]

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3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

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3 years ago