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3 years ago

a student solved the equation below by graphing log6(x-1)=log2(2x+2) which statement about the graph is true

Mathematics

2 answers:

konstantin123 [22]3 years ago

5 0

Answer:

Option A) The curves do not intercept

Anarel [89]3 years ago

3 0

The rest of the question is as following
=====================================
Which statement about the graph is true?
1. The curves do not intersect.
2. The curves intersect at one point.
3. The curves intersect at two points.
4. The curves appear to coincide.
==================================================
Solution:

See the attached figure
The blue graph represents the function log₆(x-1)
The red graph represents the function log₂(2x+2)
As shown from the figure
the curves does not meet.
Therefore, the correct statement is option 1

1. The curves do not intersect.

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Y varies jointly with x. Y=16 when x=4 and z=1/2. Find y when x=2 and z=1/4. Enter the value of y in the box. Picture attached,

bogdanovich [222]

Answer:

y = 4

Step-by-step explanation:

Given that y varies jointly with x and z then the equation relating them is

y = kxz ← k is the constant of variation

To find k use the condition y = 16 when x = 4 and z = $\frac{1}{2}$ , that is

16 = k × 4 × $\frac{1}{2}$ = 2k ( divide both sides by 2 )

8 = k

y = 8xz ← equation of variation

When x = 2 and z = $\frac{1}{4}$ , then

y = 8 × 2 × $\frac{1}{4}$ = 16 × $\frac{1}{4}$ = 4

6 0

3 years ago

A square is inscribed in a circle. If the area of the square is 9 in^2, what is the ratio of the circumference of the circle to

Alja [10]

Answer:

√2π / 4

Step-by-step explanation:

3 0

3 years ago

Me voy a comprar una barra de cortinas. Como vivo en un doceavo piso, me interesa saber cuál es la longitud máxima que puedo met

iren2701 [21]

Answer:

The maximum length of the bar I could fit in the elevator is 2.87 meters.

Step-by-step explanation:

The question in English is

I'm going to buy a curtain rod. Since I live on a twelfth floor, I'm interested in knowing the maximum length I can put in the elevator.

So I take out my pocket tape measure, which only measures up to a meter and I measure the floor which turns out to be a square of 1m x 1m, but for the height it doesn't give.

However, there's a sticker on the lift box that says it has a capacity of 2,500 litres.

With all this information, what is the maximum length of bar that would fit me in the elevator?

step 1

Find the height of the elevator

we know that

The elevator has a capacity of 2,500 litres.

$1\ m^3=1,000\ L$

$2,500\ L=2.5\ m^3$

The volume of the elevator is given by

$V=Bh$

where

B is the area of the base

h is the height of the elevator

we have

$B=(1)(1)=1\ m^2\\V=2.5\ m^3$

substitute

$2.5=(1)h\\h=2.5\ m$

step 2

Find the diagonal of the base of the elevator

Let

d ---> diagonal of the base of the elevator

Applying the Pythagorean Theorem

$d^2=b^2+b^2$

substitute

$d^2=1^2+1^2$

$d=\sqrt{2}\ m$

step 3

Find the diagonal of the rectangular prism (elevator)

Let

D ----> diagonal of the rectangular prism

d ---> diagonal of the base of elevator

h ----> height of the elevator

Applying the Pythagorean Theorem

$D^2=d^2+h^2$

substitute

$D^2=(\sqrt{2})^2+2.5^2$

$D^2=8.25\\D=2.87\ cm$

therefore

The maximum length of the bar I could fit in the elevator is 2.87 meters.

3 0

3 years ago

Solve the inequality and graph the solution set on a number line 200-4.5x less then equal to 20

rodikova [14]

ANSWER

$x \geqslant 40$

See graph

EXPLANATION

The given inequality is

$200 - 4.5x \leqslant 20$

Group like terms.

$- 4.5x \leqslant 20 - 200$

$- 4.5x \leqslant - 180$

Divide both sides by -4.5 and reverse the sign of the inequality.

$x \geqslant \frac{ - 180}{ - 4.5}$

$x \geqslant 40$

See attachment

3 0

3 years ago

What is 0.26 repeating (.26 reappearing) as a fraction in simplest form?

IRISSAK [1]

Is it 0.262626...?
repeating decimals you have to write at least two of the repeats so it is obvious what is repeating.

1000x = 262.62626...
10x = 2.62626...
1000x - 10x = 260
990x = 260
x = 26/99

6 0

4 years ago