Explain a method you could use to find the value of x.

I guessed Andy was 32 years old, but he's actually 38 years old. What is my percent of error?

Margaret [11]

Answer:

32 of 38 is 84.210% that's your answer

5 0

3 years ago

Read 2 more answers

Jacob and Ashley run an animal shelter that houses only cats and dogs. they have 37 adult animals 21 cats and 16 dogs. A group o

Bond [772]

The most likely number of cats = 21/37 x 8 = 4.54 ≈ 5 cats
The most likely number of dogs = 16/37 x 8 = 3.46 ≈ 3 dogs

The probability of that arrangement happening = 1/8^2 = 1/64 = 0.00156

3 0

3 years ago

The volume of a sphere is 9,198.11 ft3. To the nearest foot, what is the radius of the sphere? Use 3.14 for π.

deff fn [24]

Answer: 1.3 ft

Step-by-step explanation:

Using the formula

V=4

3πr3

Solving forr

r=(3V

4π)⅓=(3·9.2

4·π)⅓≈1.29978

7 0

3 years ago

the weight of abes dog can be found using the expression 2(x+3), where x is the number of weeks. the weight of karens dog can be

ivanzaharov [21]

To find this out we can set the expressions equal to each other:

$\sf 2(x+3)=3(x+1)$

Solve for 'x', first distribute:

$\sf 2x+6=3x+3$

Subtract 3 to both sides:

$\sf 2x+3=3x$

Subtract 2x to both sides:

$\boxed{\sf x=3}$

So the answer is yes, after 3 weeks the dogs will be the same weight(because 'x' represents # of weeks).

7 0

3 years ago

Solve and show steps. will award brainliest.

a_sh-v [17]

For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

Distance = Net Speed x Time of travel

For the upstream ride of the board we know that Distance is 22.5 miles and Net Speed is (S-6). Therefore, the above equation will become:

$22.5=(S-6)\times T_{1}$ where $T_{1}$ represents the time taken to travel upstream.

We can rearrange the above equation to be:

$T_{1}=\frac{22.5 }{S-6}$ ......................(Equation 1)

By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by $T_{2}$ ) will be:

$T_{2}=\frac{22.5}{S+6}$ ................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: $T_{1}+T_{2}=9$ ............(Equation 3)

Plugging in the values of $T_{1}$ and $T_{2}$ from (Equation 1) and (Equation 2) into (Equation 3), we get:

$\frac{22.5 }{S-6} +\frac{22.5 }{S+6}=9$

Simplifying the above we will get a quadratic equation:

$9S^2-45S-54=0$

The roots of this quadratic equation are:

$S=-1$ and $S=6$

Since, speed cannot be negative, $S=-1$ is out of consideration.

The speed of the boat in the lake is thus $S=6$ miles per hour.

But we have a problem with S=6 too. The problem is that if S=6, then the boat will not be able to move upstream.

Let us solve problem 2

We are given that: $\frac{x-2}{x+3}+\frac{10x}{x^2-9}$

We can rewrite it as:

$\frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)}$

$\frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)}$

Now, the numerator can be simplified as:

$\frac{x^2+10x+6}{(x-3)(x+3)} =\frac{(x+3)(x+2)}{(x-3)(x+3)} =\frac{x+2}{x-3}$

Thus, our final simplified answer is:

$\frac{x+2}{x-3}$

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is $x\neq \pm 3$

8 0

4 years ago