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4 years ago

Please help! Geometry tangents

Mathematics

1 answer:

Ymorist [56]4 years ago

7 0

Since ∠ABC is a right angle, we shall use Pythagoras Theorem to solve for x.

$\boxed {\boxed{a^2 + b^2 = c^2}}$

$x^2 + 7^2 = ( x + 5)^2$

$x^2 + 49 = x^2 + 10x + 25$

$10x = 49 - 25$

$10x = 24$

$x = 2.4$

Answer: x = 2.4

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If the point (3,3) lies on the graph of y=f(x), what point lies on y=f(x-2)?

My name is Ann [436]

Answer:

A. (5, 3)

Step-by-step explanation:

Making use of the hint:

x -2 = 3

x = 5 . . . . . . add 2

The point (x, 3) is on the graph for x=3.

The point (x-2, 3) is on the graph for x=5, so ...

the graph of (x, f(x-2)) will include the point (5, 3).

3 0

3 years ago

Read 2 more answers

Sarah lost her science book. Her school charges a lost book fee equal to 75% of the book. Sarah received a notice she owed $60 f

garri49 [273]

Cost of book = 60
Lost book = 60 * 0.75 = 45

Equation of amount of lost book (N) = 0.75 * B

Hope this helps!

4 0

4 years ago

Math help please TY :3<br><br> HWAJDSK 10 POINTS REWARDING !!

valkas [14]

I will do Point A carefully, The others I will indicate. Start with the Given Point A. Then do the translations

A(-1,2) Original Point

Reflection: about x axis:x stays the same; y becomes -y:Result(-1,-2)

T<-3,4>: x goes three left, y goes 4 up (-1 - 3, -2 + 4): Result(-4,2)

R90 CCW: Point (x,y) becomes (-y , x ) So (-4,2) becomes(-2, - 4): Result (-2, - 4)

B(4,2) Original Point

Reflection: (4, - 2)
T< (-3,4): (4-3,-2 + 4): (1 , 2)
R90 CCW: (-y,x) = (-2 , 1)

C(4, -5) Original Point

Reflection (4,5)
T<-3,4): (4 - 3, 5 + 4): (1,9)
R90, CCW (-9 , 1)

D(-1 , -5) Original Point

Reflection (-1,5)
T(<-3,4): (-1 - 3, 5 + 4): (-4,9)
R90, CCW ( - 9, - 4)

Note: CCW means Counter Clockwise

The graph on the left is the same one you have been given.

The graph on the right is the same figure after all the transformations

7 0

3 years ago

If x < 0 and y > 0, where is the point (x, y) located?

nadya68 [22]

Answer:

(x,y) is in the Quadrant II.

Step-by-step explanation:

The Quadrant II contains the x-values that are less than 0 (negative x-values), while the y-values are greater than 0 (positive y-values). Therefore, if x < 0 and y > 0, then point (x, y) must be somewhere in Quadrant II of the Cartesian Plane.

Please mark my answers as the Brainliest if my explanations were helpful :)

3 0

3 years ago

The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).

valkas [14]

Answer:

7, 12, 17...172 (34th term)

Step-by-step explanation:

4 0

3 years ago