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2 years ago

If xy x 737 = 2832 what is xy

Mathematics

1 answer:

Naily [24]2 years ago

6 0

Answer: xy = 3.84260515604

Step-by-step explanation:

So as we know xy x 737 = 2832, so theoretically if we divided we can get the value of xy:

2832/737 = 3.84260515604

There’s your answer!

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What is the missing angles

Maksim231197 [3]

Answer:

Step-by-step explanation:

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2 years ago

The force, F (Newtons), between two objects is inversely proportional to the square of the distance, d (metres), between them. T

Anarel [89]

Answer:

$D = 0.42m$

Step-by-step explanation:

Given

Force = 0.052 N when Distance = 1.6 m

Required

Find Force; when distance = 0.74 N

The question says force is inversely proportional to square of distance;

Let F represent Force and D represent Distance;

Mathematically;

$F\alpha \frac{1}{D^2}$

Convert proportion to equation

$F = \frac{k}{D^2}$

Where k is the proportionality constant; This means that k is always constant

Make k the subject of formula

$F * D^2 = \frac{k}{D^2} * D^2$

$FD^2 = k$

When F = 0.052 and D = 1.6; the value of k is as follows

$0.052 * 1.6^2 = k$

$0.13312 = k$

$k = 0.13312$

When F = 0.74;

Recall that k is always constant; so $k = 0.13312$

To solve for D, we'll make use of $F = \frac{k}{D^2}$

Substitute $k = 0.13312$ and F = 0.74;

$0.74 = \frac{0.13312}{D^2}$

Multiply both sides by D²

$0.74 * D^2 = \frac{0.13312}{D^2} *D^2$

$0.74 * D^2 = 0.13312$

Divide both sides by 0.74

$\frac{0.74 * D^2}{0.74} = \frac{0.13312}{0.74}$

$D^2 = \frac{0.13312}{0.74}$

$D^2 = 0.17989189189$

Take square roots of both sides

$D = \sqrt{0.17989189189}$

$D = 0.42413664294$

$D = 0.42m$ (Approximated)

8 0

3 years ago

Let f(x) = -5x^6√x + -7/x³√x. What would f’(x) be? If anyone could show me step-by-step, I would greatly appreciate it! I’ve wor

Illusion [34]

Answer:

$f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}$

$f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}$

Step-by-step explanation:

Rather than solving this question in a more complex method by directly using the product rule and quotient rule, it can first be considered to perform some algebraic manipulation (index laws) to simplify the expression before taking the derivative.

$\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^6\ \sqrt{x}\ +\ \frac{-7}{x^3\ \sqrt{x}}\\\\f\left(x\right)\ =\ -5x^6\cdot x^{\frac{1}{2}}\ +\ \frac{-7}{x^3\cdot x^{\frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{6\ +\ \frac{1}{2}}\ +\ \frac{-7}{x^{3\ +\ \frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ +\ \frac{-7}{x^{\frac{7}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\end{array}$

Now, the derivative of the function can be calculated simply by only using the power rule, which yields

$\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\\\\f^{\prime}\left(x\right)\ =\ \left(-5\right)\left(\frac{13}{2}\right)\left(x^{\frac{13}{2}\ -\ 1}\right)\ -\ \left(7\right)\left(-\frac{7}{2}\right)\left(x^{-\frac{7}{2}\ -\ 1}\right)\\\\f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}\\\\f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}\end{array}\\\end{large}$

8 0

2 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x) i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

3 years ago

Help with this question Its about shapes

Damm [24]

Answer:

Step-by-step explanation:

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3 years ago

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