The distance between x and 8 is less than 14

Which graph shows the solution to the equation below? log Subscript 3 Baseline (x + 3) = log Subscript 0.3 (x minus 1) On a coor

Effectus [21]

Answer:

On a coordinate plane, 2 curves intersect at (1, 1). One curve curves up and to the right from quadrant 3 into quadrant 1. The other curve curves down from quadrant 1 into quadrant 4

Step-by-step explanation:

The first function is given as:

$log_3(x+3)$

The second function is given as:

$log_{0.3}(x-1)$

First we graph both the functions.

We can see that one curves up and to the right from quadrant 3 into quadrant 1. This curve is of $log_3(x+3)$

The other curve curves down from quadrant 1 into quadrant 3

Both curves interest almost at (1,1)

See the graph attached below

Blue line represents first function

Green line represents second function

The solution lies on the Red line.

8 0

3 years ago

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Ted and Millicent write articles for an online magazine. The table gives the number of new followers the two gained in each of t

seraphim [82]

Answer:

Ted 13.9

Millicent 12.3

Ted's data is more varied

Step-by-step explanation:

6 0

3 years ago

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israel started to solve a radical equation in this way: square root of x plus 6 − 4 = x square root of x plus 6 − 4 4 = x 4 squa

agasfer [191]

Lets solve our radical equation $\sqrt{x+6}-4=x$ step by step.

Step 1 add 4 to both sides of the equation:

$\sqrt{x+6} -4+4=x +4$

$\sqrt{x} +6=x+4$

Step 2 square both sides of the equation:

$\sqrt{x+6} =x+4$

$\sqrt{x+6}^2 =(x+4)^2$

$x+6=(x+4)^2$

Step 3 expand the binomial in the right hand side:

$x+6=x^2+8x+16$

Step 4 simplify the expression:

$0=x^2+8x-x+16-6$

$x^2+7x+10=0$

Step 5 factor the expression:

$(x+2)(x+5)=0$

Step 6 solve for each factor:

$x+2=0$ or $x+5=0$

$x=-2$ or $x=-5$

Now we are going to check both solutions in the original equation to prove if they are valid:

For $x=-2$

$\sqrt{x+6}-4=x$

$\sqrt{-2+6}-4=-2$

$\sqrt{4}-4=-2$

$2-4=-2$

$-2=-2$

The solution $x=2$ is a valid solution of the rational equation $\sqrt{x+6}-4=x$ .

For $x=-5$

$\sqrt{x+6}-4=x$

$\sqrt{-5+6}-4=-5$

$\sqrt{1}-4=-5$

$1-4=-5$

$-3\neq -5$

Since -3 is not equal to -5, the solution $x=-5$ is not a valid solution of the rational equation $\sqrt{x+6}-4=x$ ; therefore, $x=-5$ is an extraneous solution of the equation.

We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.

An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.

8 0

3 years ago

A bus makes a stop at 2:30, letting off 11 people and letting on 9. The bus makes another stop ten minutes later to let off 2 mo

____ [38]

Answer:

Step-by-step explanation:

18

5 0

3 years ago

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Anybody know the answer to these

borishaifa [10]

Answer:

i would just guess bc thats me

7 0

3 years ago