Can someone help me asap

Can someone please help me?

Fittoniya [83]

Answer:

x²+4x+5

Step-by-step explanation:

i provided a picture explanation.

3 0

3 years ago

Read 2 more answers

I need to show work and Idk how for 2 or 3. The -5,-2 and 8,3 is the answers just need work shown thanks

prisoha [69]

#2
2x - 4y = -2
2x + 3y = -16
------------------subtract
-7y = 14
y = 14/-7
y = -2

2x - 4y = -2
2x - 4(-2) = -2
2x + 8 = -2
2x = -2 -8
2x = -10
x = -10/2
x = -5
answer x = -2 and y = -5

check:

2(-5)- 4(-2) = -2
-10 +8 = -2
-2 = -2.....true

2x + 3y = -16
2(-5)+3(-2) = -16
-10 + (-6) = -16
-16 = -16...true

3 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago

Which of the following exponential functions represents the graph below?

mario62 [17]

Answer:

the graph corresponds to function "D" $f(x)=3\,(\frac{1}{5} )^x$

Step-by-step explanation:

Since the graph shown corresponds to an exponential "decay" (the function decreases as we move from left to right), the base of the exponent has to be a number smaller than 1 (one). So we examine the only two options that give such (options C and D which have fractions as the base - 1/3 and 1/5 respectively)

From there, we analyze which of the two functions satisfies the crossing of the y-axis at (0,3) which is clearly shown in the graph:

We study both:

function C at x = 0 gives:

$f(x)= 5\,(\frac{1}{3})^x\\f(0)=5\,(\frac{1}{3})^0 \\f(0)=5\,(1)\\f(0)=5$