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WARRIOR [948]
3 years ago
13

Can anyone help me with this and explain it will mark brainlyiest.

Mathematics
1 answer:
pav-90 [236]3 years ago
7 0

9514 1404 393

Answer:

  x = (8/9)(4 +√7) ≈ 5.9073

Step-by-step explanation:

Maybe you want to know the value of x.

The Pythagorean theorem can be used to find the length of the bottom side of the right triangle. It is ...

  bottom = √((5x)² -(3x)²) = √(16x²) = 4x

Then the cosine of the angle at lower left is ...

  cos(α) = Adjacent/Hypotenuse

  cos(α) = 4x/5x = 4/5

Now, the law of cosines can be used to find x. Using the angle whose cosine we just found, we have ...

  (4x)² = (5x)² +8² -2(5x)(8)cos(α)

  16x² = 25x² +64 -80x(4/5)

  9x² -64x +64 = 0 . . . . . put in standard form

The quadratic formula can be used to find the solutions.

  x = (-(-64) ±√((-64)² -4(9)(64)))/(2(9))

  x = (64 ±√1792)/18 = (32 ±8√7)/9 ≈ {1.204, 5.907}

In this context, only the larger solution makes sense.

  x = (8/9)(4 +√7) ≈ 5.9073

_____

<em>Formulas used</em>

<u>Law of Cosines</u>

For sides a, b, c and angle C opposite side c, ...

  c^2=a^2+b^2-2ab\cdot\cos(C)

<u>Quadratic formula</u>

For quadratic ax²+bx+c=0, the solutions are ...

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

<u>Pythagorean theorem</u>

For legs a, b, and hypotenuse c of a right triangle, ...

  c² = a² + b²

_____

<u>Alternate solution</u>

As an alternative to using the Law of Cosines, you can define "y" to be the length of the dashed bottom leg of the triangle whose hypotenuse is 4x. Two equations can be written using y and the Pythagorean theorem for the right triangles in which y is all or part of a leg. This set of equations can be solved for x to get the same result as the one shown above.

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Find the length of side x in simplest radical form with a rational denominator.
alekssr [168]

Answer:

x = \frac{2*\sqrt{3}}{3}

Step-by-step explanation:

Reference angle = 30°

Opposite side = x

Adjacent side = 2

Apply the tan trigonometric function, thus:

tan (30) = \frac{opp}{adj}

tan (30) = \frac{x}{2}

2 * tan (30) = x

2 * \frac{1}{sqrt{3}} = x (tan 30 = 1/√3)

\frac{2}{sqrt{3}} = x

Rationalize

\frac{2* \sqrt{3}}{sqrt{3} * \sqrt{3}} = x

\frac{2*\sqrt{3}}{3} = x

4 0
3 years ago
Alex stocks up for winter. He buys 32 cans of vegetables. He pays 80 cent per can for tomatoes and 40 cents per can for corn, fo
lara31 [8.8K]
X = # of cans of tomatoes, y = # of cans of corn

x + y = 32.....y = 32 - x
0.80x + 0.40y = 18

0.80x + 0.40(32 - x) = 18
0.80x + 12.8 - 0.40x = 18
0.80x - 0.40x = 18 - 12.8
0.40x = 5.2
x = 5.2 / 0.40
x = 13 <=== 13 cans of tomatoes were bought

leaving (32 - 13) = 19 cans of corn bought
4 0
4 years ago
WILL GIVE MEDAL!! A flight of stairs is supported by two columns as shown. What is the distance from the base of the stairs to t
adell [148]
I saw the image. It formed 2 right triangles. 1 small and 1 big.
small triangle has a long leg of 4 feet and hypotenuse of 6 feet.
big triangle has a long leg of (x) and hypotenuse of 6 + 12 ft.

proportion:

4/6 = x / 18
4 * 18 = 6x
72 = 6x
72/6  = x
12 = x

The distance from the base of the stairs to the tallest column is 12 feet.

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%
Naddik [55]

We have

\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}

6 0
2 years ago
I need help!!! Pleasssse!!!!
TiliK225 [7]
What is the question
6 0
3 years ago
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