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sertanlavr [38]
3 years ago
12

In an algebraic expression what is the product of 15 and c

Mathematics
2 answers:
Morgarella [4.7K]3 years ago
8 0

Answer:

15c

Step-by-step explanation:

15 multiplyed by c gives 15c

vampirchik [111]3 years ago
4 0

Answer:

The algebraic expression is 15c.

Step-by-step explanation:

because product means multiply. so its 15c

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Use the discriminant to describe the roots of each equation. Then select the best description.
Misha Larkins [42]

Answer:

see explanation

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 : a ≠ 0, then

The nature of it's roots can be determined by the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then roots are not real

For x² - 4x + 4 = 0 ← in standard form

with a = 1, b = - 4, c = 4, then

b² - 4ac = (- 4)² - (4 × 1 × 4) = 16 - 16 = 0

Hence roots are real and equal

This can be shown by solving the equation

x² - 4x + 4 = 0

(x - 2)² = 0

(x - 2)(x - 2) = 0, hence

x - 2 = 0 or x - 2 = 0

x = 2 or x = 2 ← roots are real and equal

8 0
3 years ago
Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,
rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t ,  y_2(t) = 0   \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1 $ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t  \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0,  ⇒ dx = 0

y = 9 t,  ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$ \oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx  $

                        = 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy $

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

                            $= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$  is conservative.

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3 years ago
Rational numbers to -17/10
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Answer:

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Step-by-step explanation:

In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number. ... Moreover, any repeating or terminating decimal represents a rational number.

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