3 years ago

Solve for x 27=9^x-1

Mathematics

1 answer:

Galina-37 [17]3 years ago

4 0

27 = 9^x-1

3^3 = 9^x-1

3^3 = 3^2(x-1)

3 = 2 (x-1)

1.5 = x-1

x = 2.5

You might be interested in

What is the correct standard equation of a circle with center (5,-1) and radius 12

Charra [1.4K]

The equation for a circle is

(x-h)² + (y-k)² = r²

h = your given x

y = your given k

Let's plug everything in!

(x - 5) ² + (y - (-1)) ² = 12²

Your final equation is

(x - 5) ² + (y + 1) ² = 144

<em>Hope I helped! Comment or message me if you have any questions :) </em>

6 0

3 years ago

J is the midpoint of KL. The coordinates of J are (-8,7) and the coordinates of K are (3,4) find the coordinates of L. (show wor

musickatia [10]

Assuming linear.
if k is at (3,4) and j is at (-8,7)
set. (y1,x1) and(y2,x2)

[(y2-y1),(x2-x1)]
[(-8-3),(7-4)]
distance from k to j is (-11,3)

now take coordinates of j and distance from. k->j and add them to get coordinate for L

[(-11+(-8)),(7+3)]= (-19,10)=L

7 0

3 years ago

Triangle PQR has coordinates P(–8, 3), Q(–8, 6), and R(–3, 6). If the triangle is translated by using the rule (x, y) right-arro

omeli [17]

Answer:

Step-by-step explanation:

The rule (x, y ) → (x + 4, y - 6 )

means add 4 to the original x- coordinate and subtract 6 from the original y- coordinate, thus

P(- 8, 3 ) → P'(- 8 + 4, 3 - 6 ) → P'(- 4, - 3 )

Q(- 8, 6 ) → Q'(- 8 + 4, 6 - 6 ) → Q'(- 4, 0 )

R(- 3, 6 ) → R'(- 3 + 4, 6 - 6 ) → R'(1, 0 )

6 0