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3 years ago

I’m not sure about this question can someone help me.

Mathematics

1 answer:

Alisiya [41]3 years ago

3 0

Step-by-step explanation:

$tan \: 48 \degree = \frac{x}{61} \\ \\ 1.11 = \frac{x}{61} \\ \\ x = 1.11 \times 61 \\\\ x = 67.7 feet\:$

RP = 67.7 feet

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There are 850 Douglas fir and Ponderosa pine trees in a section of forest bought by Karamazov Logging Co. The company paid an av

weeeeeb [17]

$d+p=850\\300d+225p=217500\\\\300d+300p=255000\\300d+225p=217500\\75p=37500\\p=500\\d+(500)=850\\d=350$

350 Douglas firs and 500 Ponderosa pines.

6 0

3 years ago

Read 2 more answers

I need to find the equations for these solutions

wlad13 [49]

Answer:

hope these answers helps u out

Step-by-step explanation:

PLZZ MARK ME AS BRAINLIEST

4 0

3 years ago

Read 2 more answers

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

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

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

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

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

8 0

3 years ago

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Snowcat [4.5K]

Answer:

Following are the given series for all x:

Step-by-step explanation:

Given equation:

$\bold{J_0(x)=\sum_{n=0}^{\infty}\frac{((-1)^{n}(x^{2n}))}{(2^{2n})(n!)^2}}\\$

Let the value a so, the value of $a_n$ and the value of $a_(n+1)$ is:

$\to a_n=\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}$

$\to a_{(n+1)}=\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}$

To calculates its series we divide the above value:

$\left | \frac{a_(n+1)}{a_n}\right |= \frac{\frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2}}{\frac{(-1)^2n x^{2n}}{2^{2n}(n!)^2}}\\\\$

$= \left | \frac{(-1)^{n+1} x^{2(n+1)}}{2^{2(n+1)}((n+1))!^2} \cdot \frac {2^{2n}(n!)^2}{(-1)^2n x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)!^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |$

$= \left | \frac{ x^{2n+2}}{2^{2n+2}(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\= \left | \frac{x^{2n}\cdot x^2}{2^{2n} \cdot 2^2(n+1)^2 (n!)^2} \cdot \frac {2^{2n}(n!)^2}{x^{2n}} \right |\\\\$

$= \frac{x^2}{2^2(n+1)^2}\longrightarrow 0$ for all x

The final value of the converges series for all x.

8 0

4 years ago

What’s the GCF for 30

Vilka [71]

The greatest common factor of 30 would be 30

3 0

3 years ago