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statuscvo [17]
3 years ago
10

Quick what’s the answer

Mathematics
2 answers:
GaryK [48]3 years ago
6 0
Use the Pythagorean Theorem of a² + b² = c² to solve for x. 

x² + (x + 3)² = (√117)²
x² + x² + 6x + 9 = 117
2x² + 6x + 9 = 117
2x² + 6x + 9 - 117 = 0
2x² + 6x - 108 = 0
2x² + 18x - 12x - 108 = 0
2x(x + 9) - 12(x + 9) = 0
(2x - 12)(x + 9) = 0

x = - 9, 6

Length cannot be negative so you can't use - 9. 

x = 6. Option C is your answer. 
Lapatulllka [165]3 years ago
4 0
Use the Pythagorean Theorem:
a² + b² = c²
(x + 3)² + (x)² = (√117)²
(x² + 6x + 9) + (x²) = 117
2x² + 6x + 9 = 117
2x² + 6x - 108 = 0
2(x² + 3x - 54) = 0
2(x + 9)(x - 6) = 0
2≠0, x + 9 = 0, x - 6 = 0
            x = - 9,      x = 6
Length cannot be negative so x = -9 is not valid.

Answer: x = 6
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After eating 25 percent of the pretzels, Sonya had 42 left how many pretzels did Sonya have originally??? How do I get the Answe
madreJ [45]
42 + 25% = 52 ( 25% is 1/4)
7 0
3 years ago
Find the curl of ~V<br> ~V<br> = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k
ch4aika [34]

Given

\vec v =  f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k

the curl of \vec v is

\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k

7 0
3 years ago
What is the area of this figure
konstantin123 [22]

Answer:

108

Step-by-step explanation:

add all of the numbers together and thats what you get :)

4 0
2 years ago
a fuel mixture consists of 3 parts oil and 11 parts gasoline. the oil costs $0.75 per liter, and the gasoline costs $1.60 per li
kupik [55]

Answer:

Total selling price= $663.82

Selling price per liter= $1.69

Step-by-step explanation:

Giving the following information:

Proportion of oil and gasoline:

Oil= 3/14= 0.21

Gasoline= 11/14= 0.79

The oil costs $0.75 per liter.

The gasoline costs $1.60 per liter.

<u>First, we need to calculate the number of liters of oil and gasoline required:</u>

<u></u>

Oil= 0.21*392= 82.32 liters

Gasoline= 0.79*392= 309.68

<u>Now, the total cost to produce 392 liters:</u>

Total cost= 82.32*0.75 + 309.68*1.6

Total cost= $577.23

<u>Finally, the total and unitary selling price:</u>

Total selling price= 577.23*1.15= $663.82

Selling price per liter= 663.82 / 392= $1.69

6 0
3 years ago
Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 +
timama [110]

Looks like your function is

f(x)=\dfrac x{6x^2+1}

Rewrite it as

f(x)=\dfrac x{1-(-6x^2)}

Recall that for |x|, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

If we replace x with -6x^2, we get

f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}

By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2

Solving for x gives the interval of convergence,

|x|^2

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

3 0
3 years ago
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