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2 years ago

Find the exact value of X

Mathematics

1 answer:

Dima020 [189]2 years ago

5 0

Using Pythagorean theorem, x should be 41 :)

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What is the answer to the problem i need help with?

AveGali [126]

Answer:

Step-by-step explanation:

Recall the equation for a circle:

$(x-h)^2+(y-k)^2=r^2$ , where the center is (h,k) and the radius is r.

The given equation is:

$(x+5)^2+(y+7)^2=21^2$

Another way to write this is:

$(x-(-5))^2+(y-(-7))^2=21^2$

Thus, we can see that h=-5 and k=-7.

The center is at (-5, -7).

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3 years ago

what multiplication expression can be broken apart into (3×4) + (3×4) using the distributive property? (1-4)

jok3333 [9.3K]

3 x(4+4)= (3x4)+(3x4)
I am not completely sure, but I hope I helped!! (:

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2 years ago

Jon Kent's parents subtracted $3,500 from their adjusted gross income when they listed him as an exemption on their tax return .

NeX [460]

Answer: 3,917

Step-by-step explanation:

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2 years ago

If angle a measures 64 then what does angle b measure?

Naddik [55]

Answer:

b = 26°

Step-by-step explanation:

a ; b = complementary =>

a + b = 90° } => b = 90° - 64° = 26°

a = 64°

8 0

2 years ago

Read 2 more answers

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago