Solve for x: 0.008 - 0.02x = 0.03 Then evaluate: x^2 + x + .89 =

This is just a random question don't answer im testing something out

notsponge [240]

Answer:

I think these should help.

Step-by-step explanation:

4 0

3 years ago

HELP HOW WOULD I DO THIS? WHAT WOULD THIS BE?? HELP !! PLEASE

Ymorist [56]

You have to find the area of the four triangles which is the perimeter of the base and multiply. It by your height and your case you have to find out if it’s a slant height.

6 0

3 years ago

Which of the following is a fifth degree trinomial with a quadratic term and a negative leading

ExtremeBDS [4]

Please enter in a picture

7 0

2 years ago

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

What are the two missing sides of the triangle ??

igor_vitrenko [27]

Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.

Givens
a = x
b = x + 4
c = 20

Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2

Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20 Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0
2x^2 + 8x - 4 = 0 Divide through by 2
x^2 + 4x - 2 = 0

Use the quadratic formula
a = 1
b = 4
c = - 2

$\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} 
$
$\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} + 4(1)(2) } }{2} 
$

From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6) This is an extraneous root. No line can be minus.

x1 = + 0.4495
x2 = x + 4 = 4.4495

6 0

3 years ago