If Katie needs 1 1/3 cans of paint for each room in her house, how many cans of paint will she need to

The sides of a triangle are x , x +1 , 2 x -1 and its area is x root of 10 .Find the value of x.

VARVARA [1.3K]

Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.

Substitute the known values into the formula:

x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}

Simplify:

x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}

x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]

x√10=√[2x(x)(x-1)(1)]

x√10=√[2x²(x-1)]

x√10=√(2x³-2x²)

10x²=2x³-2x²

2x³-12x²=0

2x²(x-6)=0

2x²=0 or x-6=0

x=0 or x=6

Therefore, x=6 (you can't have a length of 0).

6 0

3 years ago

A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can

svetlana [45]

Answer:

15 days

Step-by-step explanation:

Let x be the number of days needed for B to complete the job. Then x-5 is the number of days needed for A to complete the job.

In 1 day,

A completes $\dfrac{1}{x-5}$ of all work;
B completes $\dfrac{1}{x}$ of all work.

Hence, in 1 day both A and B complete $\dfrac{1}{x-5}+\dfrac{1}{x}$ of all work. A and B working together can do a work in 6 days. Then

$6\cdot \left(\dfrac{1}{x-5}+\dfrac{1}{x}\right)=1.$

Solve this equation:

$\dfrac{6x+6x-30}{x(x-5)}=1,\\ \\12x-30=x^2-5x,\\ \\x^2-17x+30=0,\\ \\D=(-17)^2-4\cdot 30=289-120=169=13^2,\\ \\x_{1,2}=\dfrac{17\pm 13}{2}=2,\ 15.$

If $x=2,$ then $x-5=-3$ that is impossible. So, B needs 15 days to complete the work alone.

6 0

3 years ago

Cosx = 1/2 tan X = 0 solve

Reil [10]

Step-by-step explanation:

see the answer in the pic

4 0

3 years ago

Read 2 more answers

Compare how adding a number to a function and adding a number to the x-value of a function change the graphs of the functions.

love history [14]

Answer:

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I hope this pic helps :))

Step-by-step explanation:

4 0

3 years ago

The curve y = |x|/ 5 − x2 is called a bullet-nose curve. find an equation of the tangent line to this curve at the point (2, 2).

diamong [38]

We have the following curve:

$y=\frac{\left | x \right |}{5-x^2}$

So we need to find an equation of the tangent line to this curve at the point $(2,2)$ . So let's find out if this point, in fact, belongs to the curve:

 $If \ x=2 \\ \\ y=\frac{\left | 2 \right |}{5-2^2}=\frac{\left | 2 \right |}{5-4}=2$ .

We also know that:

 $\left | x \right |= \left \{ {{x \ if \ x \ \geq 0} \atop {-x \ if \ x\ \textless \ 0}} \right.$ 

Given that the point is:

 $(2,2)\ that \ is: \\ x=2\ \textgreater \ 0$

Then we will say that:

$\left | x \right |=x$

Therefore:

$y=\frac{x}{5-x^2}$

Computing the derivative:

$y=\frac{x}{5-x^2} \\ \frac{dy}{dx}= \frac{(1)(5-x^2)-(-2x)x}{(5-x^2)^2}=\frac{5+x^2}{(5-x^2)^2}$

So the derivative solved for $x=2$ is in fact the slope of the line at the point $(2,2)$ , then:

$\frac{dy}{dx} |_{x=2}=\frac{5+x^2}{(5-x^2)^2}|_{x=2}=\frac{5+2^2}{(5-2^2)^2}=9 \\ \\ \therefore m=\frac{dy}{dx} |_{x=2}=9$

Finally, the tangent line is:

$y-y_1=m(x-x_1) \\ \therefore y-2=9(x-2) \\ \therefore \boxed{y=9x-16}$

This is shown in the figure below.

8 0

4 years ago