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Ray Of Light [21]
3 years ago
15

A particular circle in the standard (x,y) coordinate

Mathematics
1 answer:
matrenka [14]3 years ago
7 0
For any circle with Cartesian equation
(x-a)^2 + (y-b)^2 = r^2,
we have that the centre of the circle is (a,b), and the radius of the circle is r.

So in the case that 
(x-5)^2 + y^2 = 38,
we essentially have that
a = 5, b = 0, r^2 = 38.

So the centre of the circle is (5,0), and the radius is \sqrt{38}.
You might be interested in
Correct answers only please!
MrRissso [65]

Answer:

23.6°

Step-by-step explanation:

In this question we have to use some trigonometry to work out angle XVW. Since we are working out angles all of the trigonometric functions will have to be to the ⁻¹. The first thing we need to identify is will formula will we use out of the following:

Sin⁻¹ = Opposite ÷ Hypotenuse

Tan⁻¹ = Opposite ÷ Adjacent

Cos⁻¹ = Adjacent ÷ Hypotenuse

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

10 = Hypotenuse

4 = Opposite

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

We know which formula to use because the length won't be in the triangle for example we will be using the Sin triangle because we don't have an adjacent. If we don't have an adjacent then the other formula's won't work.

Now we substitute in the values to find the value of angle XVW

Sin⁻¹ = Opposite ÷ Hypotenuse

Sin⁻¹ = 4 ÷ 10

The value of angle XVW is 23.57817848

6 0
2 years ago
Find the term indecent of x in the expansion of (x^2-1/x)^6
Mars2501 [29]

By the binomial theorem,

\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0   ===>   3k = 12   ===>   k = 4

which corresponds to the constant coefficient

\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}

3 0
2 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
Answer and explanation please
Gnom [1K]
For the triangle, the shaded part, both the height and base are 12, so the area of it is:

1/2 (12 x 12) = 72

And the area of the whole square is 144

But what we’re looking for is the area of the unshaded region. So to find it, we’ll need to subtract the shaded from the total. So:

144 - 72 = 72

Thus, the area of the unshaded region is 72
3 0
3 years ago
Read 2 more answers
Find the area of the semicircle
Liono4ka [1.6K]

Answer:

Area of semicircle =( π r2)÷ 2

= 1/2 × 22/7 × 11 ×11

= 11 × 17.5

= 19.25 CM2

7 0
2 years ago
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