All categories

2 years ago

Can someone please help me!? I will mark them as the Brainly est

Mathematics

2 answers:

lbvjy [14]2 years ago

8 0

The answer is 8/9 / 3 1/5= 5/18

victus00 [196]2 years ago

5 0

Answer:

Your answer is 5/18

Step by step Explanation:

8/9 × 16/5

When we divide it becomes reverse so, 

8/9 × 5/16 = 5/18.

You might be interested in

In the smoky mountains of Tennessee, the percent of moisture that falls as snow rather than rain is approximated by P = 10.0 ln

Mars2501 [29]

Answer:

A. About 4,900 ft

Step-by-step explanation:

We want h such that ...

5 = 10·ln(h) -80

8.5 = ln(h) . . . . . . . add 80, divide by 10

e^8.5 = h ≈ 4914.8 . . . . take the antilog

h ≈ 4900 . . . . feet

7 0

4 years ago

3) Smurfette is glad you're getting on with your adventure, but she wants to warn you that this last

madreJ [45]

Answer:

take x=2.515834 and r=14.31386

Step-by-step explanation:

Let us call the the lenght of the side of the square as x. The total lenght of the wire is 100. Note that if we have a square of side x, it has a perimeter of 4x. That means, that we need a wire of length 4x to build a square of side x. Note that if we take a wire of length 4x, we have that the remainder is 100-4x. We will use this to form a circle. As for the square, since we are using the wire of length 100-4x to build a circle, it must happen that the perimeter of the circle is 100-4x.

Recall that the perimeter of a circle of radius r is $2\pi r$ , so in this case , we have that $2\pi r = 100-4x$ which implies that $r =\frac{100-4x}{2\pi}$ .

Now, recall that the area of a square of side x is $x^2$ . Also, recall that the area of a circle of radius r is $\pi r^2$ . In our case, the area of the circle is given by

$\pi r^2 = \pi (\frac{100-4x}{2\pi})^2$

Also, we are given that the sum of the areas of both figures is 650. That is

$x^2 + \pi (\frac{100-4x}{2\pi})^2 = x^2+\frac{(100-4x)^2}{4\pi}=650$

By substracting 650 on both sides and multiplying by $4\pi$ this equation is equivalent to

$4\pi x^2 + (100-4x)^2-650\cdot 4\pi =0$

if we expand $(100-4x)^2$ and then group each term of x, x^2 we get :

$(4\pi+16)x^2-800x+100^2-650\cdot 4\pi =0$

Recall that given a cuadractic equation of the form $ax^2+bx+c=0$ the solution is given by

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In this case, the existence of a real solution is given by the expression $b^2-4ac$ . For it to have a real solution, it must happen that $b^2-4ac\geq 0$

In this case, b=-800, c = 100^2-650\cdot 4\pi, a = 4\pi+16[/tex]

For this values, we get that $b^2-4ac = 430681\geq 0$ , so the is a solution for our problem.

In this case,

$x = \frac{-b - \sqrt[]{b^2-4ac}}{2a}= 2.515834$

and $x = \frac{-b + \sqrt[]{b^2-4ac}}{2a}= 25.4891$ are the two possible solutions for this problem. We can check that this values of x fulfill our restrictions. Note that if x=25.4891, by replacing in the value of r, we get that r=-0.3113853. Since r is the radius of the circle, it must be positive. Hence, x=25.4891 is discarted.For x=2.515834 we get a value of r=14.31386, so this is the answer