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Alexandra [31]
2 years ago
11

The parent function, f(x) = 5^x, has been vertically compressed by a factor of one-half, shifted to the left three units and dow

n two unlts.
Choose the correct function to represent the transformation.

Mathematics
1 answer:
Alona [7]2 years ago
8 0

Answer: Choice C

\displaystyle g(x) = \left(\frac{1}{2}\right)5^{x+3}-2

=====================================================

Explanation:

The 1/2 out front handles the vertical compression by 1/2. For example, if two points are vertically spaced by 10 units, then their new vertical distance is now (1/2)*10 = 5 units.

The x+3 in the exponent means "shift 3 units to the left". Effectively what's going on is that the old input x is now x+3, ie 3 units larger than before. This shifts the xy axis itself 3 units to the right. If we held the curve fixed in place while the xy axis moved like this, then it gives the illusion the curve is moving 3 units to the left.

Then finally the -2 at the very end shifts the curve down 2 units. This is because whatever the y coordinate is, we subtract 2 from it to do this vertical shifting. For example, the point (0,62.5) shifts 2 units down to (0,60.5)

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Answer:

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Find the area of the blue area if each<br> square represents 1.5 square meters.
Anni [7]

Check the picture below.

so the picture has a rectangle that is 8 units high and 12 units wide, and it has a couple of "empty" trapezoids, with a height of 5 and "bases" of 9 and 3.

now, if we just take the whole area of the rectangle and then subtract the area of those two trapezoids, what's leftover is the blue area.

\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=9\\ b=3 \end{cases}\implies \begin{array}{llll} A=\cfrac{5(9+3)}{2}\implies A=30 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Areas}}{\stackrel{rectangle}{(12\cdot 8)}~~ -~~\stackrel{\textit{two trapezoids}}{2(30)}}\implies 96-60\implies 36

3 0
2 years ago
Parallel line to 2x+3y=6 passing through (6,-3)
MissTica

Answer:

y = (-2/3)x + 1

Step-by-step explanation:

3y = -2x + 6

y = (-2/3)x + 2

y + 3 = (-2/3)(x - 6)

y + 3 = (-2/3)x + 4

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3 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.

5 0
3 years ago
Please help it’s due in 20 minutes and it’s easy i just don’t get it :( i’ll mark as brainiest !!
sukhopar [10]

Answer:

d is 18, a is 11, c is 9 and b is 12

Step-by-step explanation:

i think its this, although im only really sure for d.

im 50 percent sure with the others as they add to 50

6 0
3 years ago
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