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3 years ago

Math question. Please help

Mathematics

1 answer:

yanalaym [24]3 years ago

3 0

To find the perimeter of the triangle, you need to know the length of the hypotenuse (h). Use the Pythagorean theorem to find it.
.. h^2 = (4w)^2 + (15/2*w)^2
.. = w^2*(4^2 +(15/2)^2)
.. = w^2*(16 +56 1/4)
.. = w^2*(72 1/4)
Taking the square root, we find the hypotenuse to be
.. h = √(w^2*72.25) = 8.5w = (17/2)w

Now, the length of the triangle's perimeter is the sum of the lengths of the sides.
.. P = 4w +(15/2)w +(17/2)w
.. P = 20w

The area of the triangle is half the product of base and height:
.. A = (1/2)(4w)(15/2*w)
.. A = 15w^2

So, the ratio of perimeter to area is
.. P/A = (20w)/(15w^2) = 4/(3w) . . . . . . . . simplified expression for P/A

The problem is asking you to find (and graph) the values of w such that
.. P/A < 1
.. 4/(3w) < 1 . . . . . . substitute our value of P/A
.. 4/3 < w . . . . . . . . multiply by w

This is graphed with an open circle around the point 1 1/3, and a solid arrow to the right of that circle. The open circle shows that w=4/3 is NOT a solution, but all values greater than that are.

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Point r is at 2,1.2 and point t is at 2,2.5 on a coordinate grid. The distance between the two points is. .

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Answer:

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Step-by-step explanation:

The two points lie on the same vertical line (x=2), so the distance between them is the difference of their y-coordinates:

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Marisa bought four bottles of water.Each bottle of water was 95 cents. Write an equation with the same product as the total cost

meriva

Answer:

$5\times 0.76= 3.8$

Step-by-step explanation:

Given: Marisa bought four bottles

Price of each bottle is $0.95.

Solving with the factors as given to find the total cost paid by Marisa.

Total cost= $4\ bottles\times \$ 0.95= \$ 3.8$

∴ Total cost paid by Marisa is $3.8.

As given to write an equation with the same product as the total cost but different factors.

∴ Lets assume the number of bottles bought by Marisa be "5" and cost of each bottle be "x"

We already find the total cost she paid is $3.8.

Now, writing an equation with different factors, as total cost remain same.

$5\times x= 3.8$

Dividing both side by 5

we get x= $ 0.76

∴ The required equation is $5\times 0.76= 3.8$

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3 years ago

Please I need help!

Artemon [7]

Answer:

I think it’s 20 ft

Step-by-step explanation:

7 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago