How do I figure out 8 meters increased by 25 percent ?

What is the area, in square centimeters, of the trapezoid below? 8.5 cm 7.5 cm 15.7 cm

Contact [7]

Answer:

90.75 cm^2

Step-by-step explanation:

The area of a trapezoid is given by

A = 1/2 ( b1+b2) *h

where b1 and b2 are the lengths of the bases and h is the height

A = 1/2(8.5+ 15.7) * 7.5

1/2 (24.2) 7.5

90.75 cm^2

6 0

1 year ago

For the love of God help me !! I'm desperate for it tomorrow

Eduardwww [97]

Try to relax. Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before. But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this: (2)^a negative power = (1/2)^the same power but positive.

So:
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was: log₂ of(N) = - log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract log₂(x) from each side:

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract log(base 1/2) (x-2) from each side:

log₂(x-1) - log₂(x) = - log(base 1/2) (x-2) Notice the negative on the right.

The left side is the same as log₂[ (x-1)/x ]

==> The right side is the same as +log₂(x-2)

Now you have: log₂[ (x-1)/x ] = +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' : x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that x=2.618 is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.

There,now. Doesn't that feel better.

4 0

2 years ago

Mr. James has cylindrical beakers that measure 4 inches in diameter and 9 inches high. What is the volume contained within the b

goblinko [34]

So 4 inches diameter result the radius is 4/2 = 2 inches

area of base = pir^2 = 3,14*2^2 = 3,14*4 = 12,56 inches squared

volume = area of base *height

V = 12,56 *9 = 113,04 so rounded 113,05 inches cubed

hope this will help you

3 0

3 years ago

Find the pattern: 1, 1/4, 1/9, 1/16, 1/25

alexandr1967 [171]

The sequence is not geometric or arithmetic because there is no common difference or common ratio between each term.
Not a Geometric or Arithmetic Sequence

8 0

2 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for h:

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for h.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

7 0

2 years ago