All categories

3 years ago

18. A circle in the standard (x,y) coordinate plane is

Mathematics

1 answer:

Lana71 [14]3 years ago

3 0

9514 1404 393

Answer:

Step-by-step explanation:

If the circle is tangent to the x-axis at 4, the center lies on the line x=4.

If the circle is tangent to the y-axis at 4, the center lies on the line y=4.

If the center of the circle is (x, y) = (4, 4) and it is tangent to the axes, then the radius is 4.

The standard-form equation of the circle centered at (h, k) with radius r is ...

(x -h)² +(y -k)² = r²

For the values (h, k) = (4, 4) and r = 4, the equation is ...

(x -4)² +(y -4)² = 16 . . . . . . matches choice F

You might be interested in

At the beginning of the year, Jason had $80 his savings account. Each month, he added $15 to his account. Write an expression fo

Tems11 [23]

Expression: 80+15x=y
(x means months)
so at the end of the year jason will have $260

8 0

3 years ago

Read 2 more answers

C is directly proportional to d such that

defon

Answer:

we have

C = 1.5d

d=C/1.5

Step-by-step explanation:

when d is 12

C=1.5×12=18

when c=15

d=15/1.5=10

5 0

3 years ago

The radius of circle a is 6 in. the radius of circle b is 5 in. greater than the radius of circle a. the radius of circle c is 5

Agata [3.3K]

Answer:

Step-by-step explanation:

5 0

1 year ago

I need help with this

cupoosta [38]

Answer:

x=45

Step-by-step explanation:

This is a triangle so it interior degrees will add up to 180.

Since Line AC is tangent to Circle O at point c, Point C measures 90 degrees.

We can find OAC by using the triangle interior theorem,

$45 + 90 + x = 180$

$135 + x = 180$

$x = 45$

3 0

2 years ago

Can anyone help me solve a trigonomic identity problem and also help me how to do it step by step?

dusya [7]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad csc(\theta)=\cfrac{1}{sin(\theta)}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\$

$\bf \cfrac{cos(\theta )cot(\theta )}{1-sin(\theta )}-1=csc(\theta )\\\\
-------------------------------\\\\
\cfrac{cos(\theta )\cdot \frac{cos(\theta )}{sin(\theta )}}{1-sin(\theta )}-1\implies \cfrac{\frac{cos^2(\theta )}{sin(\theta )}}{\frac{1-sin(\theta )}{1}}-1\implies 
\cfrac{cos^2(\theta )}{sin(\theta )}\cdot \cfrac{1}{1-sin(\theta )}-1
\\\\\\
\cfrac{cos^2(\theta )}{sin(\theta )[1-sin(\theta )]}-1\implies 
\cfrac{cos^2(\theta )-1[sin(\theta )[1-sin(\theta )]]}{sin(\theta )[1-sin(\theta )]}$

$\bf \cfrac{cos^2(\theta )-1[sin(\theta )-sin^2(\theta )]}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{cos^2(\theta )-sin(\theta )+sin^2(\theta )}{sin(\theta )[1-sin(\theta )]}
\\\\\\
\cfrac{cos^2(\theta )+sin^2(\theta )-sin(\theta )}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{\underline{1-sin(\theta )}}{sin(\theta )\underline{[1-sin(\theta )]}}
\\\\\\
\cfrac{1}{sin(\theta )}\implies csc(\theta )$

7 0

3 years ago