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2 years ago

What is 122 / 600?

600" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

kipiarov [429]2 years ago

5 0

The answer : 0.20333333333

dmitriy555 [2]2 years ago

4 0

Answer:

122÷600

0.203333334

that's my answer please

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3x^2+5x_=2+_

tatiyna

IT DOES NOT MATTER YOU WILL NOT BE ABLE TO SOLVE THIS QUESTION
YOU WOULD GET 5x = 2 + 3x^2 AND YOU WOULD STILL NOT BE ABLE TO SOLVE YOU COULD GET 3x^2 = 2 + 5x AND YOU COULD NOT SOLVE.
THE REASON WHY IS BECAUSE YOU HAVE TO HAVE THE SAME X TO ADD THEM OR MINUS THEM< WHICH IN THIS CASE ONE IS REGULAR BUT THE OTHER IS SQUARED. AND ALSO THE TWO NEEDS SOMEONE TO GO WITH< BUT THE IS ONLY ONE OF HIS TYPE SO HE CAN'T.
WELCOME :/

3 0

3 years ago

I need help, urgently.

Korvikt [17]

Answer:

y = -0.5(x - 4)^2 + 5.

y = -0.5x^2 + 4x - 3.

Step-by-step explanation:

The vertex is at (4, 5). so we have:

f(x) = a(x - 4)^2 + 5

When x = 0 y = -3 so substituting in the above:

- 3 = a(0-4)^2 + 5

-8 = 16a

a = -0.5.

So the vertex form is y = -0.5(x - 4)^2 + 5.

Standard form:

y = -0.5(x^2 - 8x + 16) + 5

y = -0.5x^2 + 4x - 8 + 5

y = -0.5x^2 + 4x - 3.

5 0

3 years ago

HELP!! ASAP PLEASE!! WILL GIVE BRAINLIEST!!

patriot [66]

Answer:

6^-4 ÷ 3^-4

6^-4/3^-4

Since their powers are negative

Flip them both so the negative index is lost.

It now becomes

3^4/6^4

81/1296

=1/16

6 0

3 years ago

Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the

sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is $\mathbf{ a^2 ( \pi -2)}$

Step-by-step explanation:

The surface area of the sphere is:

$\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA$

and the cylinder $x^2 + y^2 =ax$ can be written as:

$r^2 = arcos \theta$

$r = a cos \theta$

where;

D = domain of integration which spans between $\{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}$

and;

the part of the sphere:

$x^2 + y^2 + z^2 = a^2$

making z the subject of the formula, then :

$z = \sqrt{a^2 - (x^2 +y^2)}$

Thus,

$\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}$

Similarly;

$\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}$

So;

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}$ $\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}$

From cylindrical coordinates; we have:

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}$

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

$A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA$

$A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta$ $A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta$