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3 years ago

I need help on this question

Mathematics

2 answers:

zvonat [6]3 years ago

5 0

Answer:

x=32

Step-by-step explanation:

klasskru [66]3 years ago

3 0

Answer:

x= 32

Step-by-step explanation:

Remember all triangles sum up to 180 degrees.

Equation= 98+50+x=180

Step 1- Add to simplify

148+x= 180

Step 2- Subtract 148 to both sides.

148+x= 180

-148 148

x= 32

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What is the simplified form of x plus 4 over x squared minus 3x minus 10⋅x minus 3 over x squared plus x minus 12? (6 points)

melomori [17]

Answer:

The simplified form is 1 over the quantity x plus 2 times the quantity

x minus 5 ⇒ 1/(x+2)(x-5) ⇒ last answer

Step-by-step explanation:

* Lets write the product of the two fraction

∵ $\frac{x+4}{x^{2}-3x-10} *\frac{x-3}{x^{2}+x-12}$

- At first factorize the denominators

# x² - 3x - 10

∵ x² = x × x ⇒ 1st term in the 1st bracket and 1st term in the

2nd bracket

∵ -10 = 2 × -5 ⇒ 2nd term in the 1st bracket and 2nd term in the

2nd bracket

∵ x × - 5 = -5x ⇒ ext-reams

∵ x × 2 = 2x ⇒ means

∵ 2x + -5x = -3x ⇒ middle term

∴ x² - 3x - 10 = (x + 2)(x - 5)

# x² + x - 12

∵ x² = x × x ⇒ 1st term in the 1st bracket and 1st term in the

2nd bracket

∵ -12 = -3 × 4 ⇒ 2nd term in the 1st bracket and 2nd term in the

2nd bracket

∵ x × 4 = 4x ⇒ ext-reams

∵ x × -3 = -3x ⇒ means

∵ 4x + -3x = x ⇒ middle term

∴ x² + x - 12 = (x - 3)(x + 4)

* Lets write the fractions after factorization

∴ $\frac{x+4}{(x+2)(x-5)}*\frac{x-3}{(x-3)(x+4)}$

- lets simplify the fractions by cancel (x - 3) up with (x - 3) down

and cancel(x + 4) up with (x + 4) down

∴ $\frac{1}{(x+2)(x-5)}*1=\frac{1}{(x+2)(x-5)}$

* The simplified form is 1 over the quantity x plus 2 times the

quantity x minus 5

3 0

3 years ago

50,000 is 20% of what?

Nimfa-mama [501]

Answer:

250000

Step-by-step explanation:

20% of x is 50000

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

50000=20x/100

50000= x/5

5(50000) = x

250000 = x

5 0

3 years ago

Read 2 more answers

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:

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

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

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

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