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UkoKoshka [18]
3 years ago
10

Help please how do I find x and y

Mathematics
1 answer:
Dafna1 [17]3 years ago
7 0

Answer:

Option 3

Step-by-step explanation:

tan33°=x/7

x=4.5

4.5^2+7^2=y^2

20.25+49=y^2

y=√69.25

y=8.3

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10 points please help and explain how ;) brainliest answer first person to get it right
OverLord2011 [107]

Answer:

Open dot at -1 and shaded to the right.

Step-by-step explanation:

First you have to solve the inequality are you get |y|>-1

Now you can plot the inequality by placing an open dot on -1 since the inequality is greater than.

Since the inequality is greater than, you also have to shade when all y values are greater than -1 so you would shade to the right.

5 0
3 years ago
The graph below represents the price of sending video messages using the services of a phone company. What is the constant of pr
valina [46]

Answer:

I believe your answer is 24

Step-by-step explanation:24/1

4 0
3 years ago
Please help me with this question ​
gayaneshka [121]

9514 1404 393

Answer:

  D

Step-by-step explanation:

The leading term is of odd degree and negative coefficient. The odd degree tells you the end behaviors will have opposite signs. The negative coefficient tells you the left end behavior will be positive, and the right end behavior will be negative.

  \textbf{D.  }\lim\limits_{x\to -\infty}f(x)=\infty\quad\lim\limits_{x\to \infty}f(x)=-\infty

5 0
3 years ago
SHOW YOUR WORK FOR ANOTHER BRAINLIEST ;D
omeli [17]

Answer: 1,500 x 12

Step-by-step explanation:

They make 1500 a month, there are 12 months in a year.

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4 years ago
Read 2 more answers
Write an equation of the circle with center , −45 and radius 6 .
IceJOKER [234]
The general equation for a circle, x^2+y^2=1, falls out of the Pythagorean Theorem, which states that the square of the hypotenuse of a right triangle is always equal to the sum of the squares of its legs (you might have seen this fact written like a^2+b^2=c^2, where <em>a </em>and <em>b</em> are the legs of a right triangle and <em>c </em>is its hypotenuse. When we fix <em /><em>c</em> in place and let <em>a </em>and <em>b </em>vary (in a sense, at least; their values are still dependent on <em>c</em>), the shape swept out by all of those possible triangles is a circle - a shape defined by having all of its points equidistant from some center.

How do we modify this equation to shift the circle and change its radius, then? Well, if we want to change the radius, we simply have to change the hypotenuse of the triangle that's sweeping out the circle in the first place. The default for a circle is 1, but we're looking for a radius of 6, so our equation, in line with Pythagorus's, would look like x^2+y^2=6^2, or x^{2} +y^2=36.

Shifting the center of the circle is a bit of a longer story, but - at first counterintuitively - you can move a circle's center to the point (a,b) by altering the x and y portions of the equation to read:

(x-a)^2+(y-b)^2
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3 years ago
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