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MrRissso [65]
3 years ago
10

Evaluate the algebraic expression for the given value(s) of the variable(s).

Mathematics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

-18

Step-by-step explanation:

2x²-7xy+3y²

2(3²)-7(3)(4)+3(4²)

2(9)-84+3(16)

18-84+48

18+48-84

-18

You might be interested in
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
Helppppppppppp:)))))))))
Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}

1. What is the domain?

  • Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.

Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>c</u><u>t</u><u>u</u><u>a</u><u>l</u><u> </u><u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>0</u><u>,</u><u>2</u><u>,</u><u>3</u><u>}</u>

2. What is the range?

  • Because domain is set of all x-values. Then what do you think the range is? That's right! The range is <u>s</u><u>e</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>l</u><u>l</u><u> </u><u>y</u><u>-</u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>s</u><u>.</u> If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!

Therefore, gather y-values from the ordered pairs. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>-</u><u>2</u><u>,</u><u>-</u><u>1</u><u>,</u><u>1</u><u>,</u><u>2</u><u>}</u>

3. Is the relation a function?

  • All functions are relations but not all relations are functions. Function is a set of ordered pairs where <u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>n</u><u>o</u><u>t</u><u> </u><u>r</u><u>e</u><u>p</u><u>e</u><u>t</u><u>i</u><u>t</u><u>i</u><u>v</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>i</u><u>n</u><u> </u><u>a</u><u> </u><u>s</u><u>e</u><u>t</u><u>,</u><u> </u><u>t</u><u>h</u><u>e</u><u>r</u><u>e</u><u> </u><u>c</u><u>a</u><u>n</u><u>n</u><u>o</u><u>t</u><u> </u><u>b</u><u>e</u><u> </u><u>m</u><u>o</u><u>r</u><u>e</u><u> </u><u>t</u><u>h</u><u>a</u><u>n</u><u> </u><u>o</u><u>n</u><u>e</u><u> </u><u>s</u><u>a</u><u>m</u><u>e</u><u> </u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>.</u> Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.

Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.

These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.

<em>F</em><em>r</em><em>o</em><em>m</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>Q</em><em>1</em><em>'</em><em>s</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>,</em><em> </em><em>t</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>t</em><em>w</em><em>o</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em>s</em><em>,</em><em> </em><em>p</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>c</em><em>h</em><em>o</em><em>o</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em> </em><em>t</em><em>o</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>u</em><em>n</em><em>d</em><em>e</em><em>r</em><em>l</em><em>i</em><em>n</em><em>e</em><em>)</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>n</em><em>o</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>f</em><em>i</em><em>r</em><em>s</em><em>t</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>s</em><em>a</em><em>m</em><em>e</em><em> </em><em>2</em><em>'</em><em>s</em><em>)</em><em>.</em><em> </em>

Good luck on your assignment, have a nice day!

4 0
3 years ago
Need awnsers please help
sleet_krkn [62]
The line of reflection is what the graph flips over. You can find the line with two points, and a point on the reflection line is the midpoint of a point and the corresponding point in the after-image.
The first one reflects over the y-axis, or x=0. One point is (-2, 1) and its corresponding point is (2, -1). The midpoint is found by the average of the two coordinates, which is (0,0). Pick another pair of points and find the midpoint which you should get (x,0).
You have two points (0,0) and (x,0) and they form a line, which is the y-axis, or x=0.
The line of reflection for the 1st one is x=0 (y-axis).
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3 years ago
Write 0.511 as a percent
Kazeer [188]

51.1 % is the answer.

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3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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3 years ago
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