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3 years ago

Pls help ASAP help pls

Mathematics

1 answer:

Savatey [412]3 years ago

4 0

Answer:

Angle Bisector

Step-by-step explanation:

The top angle is split into two equal sections

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Given PQ = y, RS = 4y - 15, QR = 2 +6, and PS = 4a - 6, solve for the length of RS and QR.

Anna11 [10]

Answer:

Check pdf

Step-by-step explanation:

Download pdf

3 0

3 years ago

My twin and I constantly argue about who is the elder. He claims that just because he was born first, he is older than I am. I r

NemiM [27]

Time zones and time offsets. A time zone is a geographical region in which residents observe the same standard time. A time offset is an amount of time subtracted from or added to Coordinated Universal Time (UTC) time to get the current civil time, whether it is standard time or daylight saving time (DST)

I hope my answer has come to your help. God bless you and have a nice day ahead!

5 0

3 years ago

What is the equation of the quadratic graph with a focus of (3, 6) and a directrix of y = 4? f(x) = one fourth (x − 3)2 + 1 f(x)

mina [271]

Let point (x, y) be any point on the graph, than the distance between (x, y) and the focus (3, 6) is sqrt((x - 3)^2 + (y - 6)^2) and the distance between (x, y) and the directrix, y = 4 is |y - 4|

Thus sqrt((x - 3)^2 + (y - 6)^2) = |y - 4|
(x - 3)^2 + (y - 6)^2 = (y - 4)^2
x^2 - 6x + 9 + y^2 - 12y + 36 = y^2 - 8y + 16
x^2 - 6x + 29 = -8y + 12y = 4y
(x - 3)^2 + 20 = 4y
y = 1/4(x - 3)^2 + 5

Required answer is f(x) = one fourth (x - 3)^2 + 5

8 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

Select all the correct graphs. Choose the graphs that indicate equations with no solution.

Natali [406]

Answer:

The first and last graph.

General Formulas and Concepts:

Algebra I

Solving systems of equations graphically

Step-by-step explanation:

In order for a systems of equations to have a solution set, the 2 graphs must intersect at at least 1 point. Here, we see that graphs 1 and 5 do not intersect each other at all.

Therefore, the rest of the graphs have solutions and #1 and #5 do no have any solutions.

3 0

3 years ago