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solmaris
2 months ago
3

How many pairs of Whole numbers have a sum of 40

Mathematics
2 answers:
AnnZ [12.3K]2 months ago
6 0
Whole numbers include counting numbers starting from zero {0, 1, 2, 3, 4, ...} and are all positive.
If pairs of whole numbers sum to 40, the total number of pairs is 41 since each time you decrease one number by 1 and increase the other by 1, the sum remains 40.

Examples:
40 + 0 = 40
39 + 1 = 40
38 + 2 = 40
...
lawyer [12.5K]2 months ago
5 0
Prime numbers up to 40 are:

2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37.

There exist three prime pairs whose sums equal 40: {3, 37}, {11, 29}, and {17, 23}.
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The side length, s, of a cube is x – 2y. If V = s3, what is the volume of the cube? x3 – 6x2y + 12xy2 – 8y3 x3 + 6x2y + 12xy2 +
Leona [12618]

V = x³ - 6x²y + 12xy² - 8y³

V = (x - 2y)³

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   = x³ - 4x²y + 4xy² - 2x²y + 8xy² - 8y³ ( combine similar terms )

  = x³ - 6x²y + 12xy² - 8y³




8 0
18 days ago
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A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds.
lawyer [12517]

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

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a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>
7 0
1 month ago
mridul jogs around a rectangular park of 100m by 80m at the rate of 9km per hour. how much time will he take in jogging 6 rounds
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Answer:

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1 month ago
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