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3 years ago

Type the correct answer in each box. If necessary, use / for the fraction bar(s).

Mathematics

2 answers:

Tom [10]3 years ago

8 0

Answer:

x=5/6

y=8/5

That's the answer, I got it right.

xz_007 [3.2K]3 years ago

6 0

The value of x is 5/6 and y is 8/5.

Step-by-step explanation:

Given,

12x+15y=34 Eqn 1

-6x+5y=3 Eqn 2

Multiplying Eqn 2 by 2

$2(-6x+5y=3)\\-12x+10y=6\ \ \ Eqn 3$

Adding Eqn 1 and 3

$(12x+15y)+(-12x+10y)=34+6\\12x+15y-12x+10y=40\\25y=40$

Dividing both sides by 25

$\frac{25y}{25}=\frac{40}{25}\\y=\frac{8}{5}$

Putting y= 8/5 in Eqn 1

$12x+15(\frac{8}{5})=34\\12x+3*8=34\\12x+24=34\\12x=34-24\\12x=10$

Dividing both sides by 12

$\frac{12x}{12}=\frac{10}{12}\\x=\frac{5}{6}$

The value of x is 5/6 and y is 8/5.

Keywords: Linear equations, fractions

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#LearnwithBrainly

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Answer:

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Find the equation of the sphere in standard form centered at (−6,10,5) with radius 5.

Dovator [93]

Answer:

The equation of the sphere in standard form is

$x^{2} +y^{2}+z^{2} +12 x-20 y-10 z+136=0$

Step-by-step explanation:

<u>Step 1</u>:-

The equation of the sphere having center and radius is

$(x-h)^{2} +(y-k)^{2} +(z-l)^{2} = r^{2}$

Given centered of the sphere is (-6,10,5) and radius r=5

$(x-(-6))^{2} +(y-10)^{2} +(z-5)^{2} = 5^{2}$

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If sin(x) = 5/13, and x is in quadrant 1, then tan(x/2) equals what?

Rufina [12.5K]

$x$ is in quadrant I, so $0$ , which means $0$ , so $\dfrac x2$ belongs to the same quadrant.

Now,

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

Since $\sin x=\dfrac5{13}$ , it follows that

$\cos^2x=1-\sin^2x\implies \cos x=\pm\sqrt{1-\left(\dfrac5{13}\right)^2}=\pm\dfrac{12}{13}$

Since $x$ belongs to the first quadrant, you take the positive root ( $\cos x>0$ for $x$ in quadrant I). Then

$\tan\dfrac x2=\pm\sqrt{\dfrac{1-\frac{12}{13}}{1+\frac{12}{13}}}$

$\tan x$ is also positive for $x$ in quadrant I, so you take the positive root again. You're left with

$\tan\dfrac x2=\dfrac15$

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3 years ago