How to solve - (12/h) -7f h=3 f=7

Use the given information to write an equation of the circle. center at (-7,-3), radius 2

Igoryamba

The equation of a circle is written as (x-h)^2 +(y-k)^2 = r^2

h and K are the center points of the circle and r is the radius.

replace the letters with the provided center and radius:

(x- (-7))^2 + (y- (-3))^2 = 2^2

Simplify:

(x+7)^2 + (y+3)^2 = 4

7 0

3 years ago

Julie has a large red apple in her hand that is 4 ft off the ground. A blue bird sees the apple at an

Dafna1 [17]

Answer:

Step-by-step explanation:

so from the silly drawing, you can see that the triangle is 4 feet off the ground, and that we want to find the Opp side from our angle of 15 °

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

use TOA to find the Opp side of the triangle .. which is labeled X

Tan(15) = X / 15

15* Tan(15) = X

4.019 = X

so the bird is up 4 feet in the tree.. not very high up, plus the 4 feet that the apple is already off the ground, soo

the tree... to where the bird is, at least, is 8.019 feet..

did this make sense? :/

hope the drawing helped :P

6 0

2 years ago

50 POINTS!!!

Goryan [66]

9514 1404 393

Answer:

f(x) = x
g(x) = -2x+1
f(x) -(-g(x)) = -x+1
f(x) +g(x) = -x+1
f(x)-(-g(x)) = (f+g)(x) is true for all functions f and g, linear or not

Step-by-step explanation:

We can define a couple of linear functions as ...

f(x) = x

g(x) = -2x+1

Then the reflected function -g(x) is ...

-g(x) = -(-2x +1) = 2x -1

And the difference from f(x) is ...

f(x) -(-g(x)) = x -(2x -1) = -x +1 . . . . f(x) -(-g(x))

We want to compare that to the sum of the functions:

f(x) +g(x) = x +(-2x +1) = -x +1 . . . . f(x) +g(x)

The two versions of the function expression have the same value.

These results are a property of addition, so do not depend on the nature of f(x) or g(x). They will hold for every function.

3 0

2 years ago

For the school play, the advance tickets cost $3, while tickets at the door cost $5. Thirty more tickets were sold at door in ad

Talja [164]

Answer:

310 tickets were sold in advance before school pay and 340 door tickets were sold in advance.

Step-by-step explanation:

Given:

Let number of tickets sold in advance be 'x'.

Cost of advance ticket = $3

Cost of door tickets = $5

Also given:

Thirty more tickets were sold at door in advance

Hence number of door tickets sold = $x+30$

Total Money Collected = $2630

Now we can say that Total Money Collected is equal to sum of number of tickets sold in advance multiplied by Cost of advance ticket and number of door tickets sold multiplied by Cost of door tickets.

Framing in equation form we get;

$3x+(x+30)5=2630$

Solving the equation to find the value of x we get;

$3x+5x+150=2630$

Combining the like terms we get;

$3x+5x=2630-150\\\\8x= 2480$

Now Dividing 8 on both side using division property we get;

$\frac{8x}{8} =\frac{2480}{8}\\ \\x=310$

Substituting the value of x to find number of door tickets been sold.

number of door tickets sold = $x+30=310+30 =340$

Hence, 310 tickets were sold in advance before school pay and 340 door tickets were sold in advance.

8 0

2 years ago

Which shows all the exact solutions of 2sec^2 (x)-tan^4 (x)=-1? Give your answer in radians.

shutvik [7]

Answer:

a. 1/3pi+kpi and 2/3pi+kpi

Step-by-step explanation:

2Sec²x

=2+2tan²x-(tan²x)²=-1

Let tan²x=u

-u²+2u+2=-1

-u²+2u+3=0

u²-2u-3=0

(u+1)(u-3)=0

u=-1,u=3

(tan x)²=-1(No solution)

(tan x)²=3

>>tan x=±sqrt(3)

I) tan x=sqrt(3)

>>tan(pi/3)=sqrt 3

Tan(pi+A) =tanA

Tan(pi+pi/3)=tan(pi/3)

>>tan(4/3pi)=sqrt(3)

>>x=4/3pi and 1/3pi

General solution: 1/3pi+kpi

ii) tan x=-sqrt(3)

x=2/3pi(At 2nd quadrant)

x=5/3pi(at 4th quadrant)

General solution: 2/3pi+kpi

4 0

1 year ago