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4 years ago

6

In this lesson, you learned new formulas to find the area of a triangle. How are these formulas different from the commonly know

n A = ½bh? What would be the use of each formula? Would you use radians in any of the formulas? Give examples.

1 answer:

Yuki888 [10]4 years ago

3 0

Aside from the conventional formula for triangle, A=<span>½bh which is only applicable to problems where the base and height are already given and the triangle is a right triangle having a degree of 90. There are some formulas in getting the area of a triangle:

>Given three sides of the triangle, use Heron's Formula
A= sqrt(s(s-a)(s-b)(s-c))
s= (a+b+c)/2
>Given two sides with an included angle
</span>Area = <span>1/2 </span><span>ab sin (tetha)
</span><span>tethat should be in degrees

</span>

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Based on the diagram below, what is the measure of arc DC ?

emmainna [20.7K]

Given:

A diagram of a circle O, $arc(AB)=y^\circ, arc(CD)=(80+y)^\circ$ .

To find:

The measure of arc DC.

Solution:

According to the angle of intersecting chords theorem:

Angle at intersection of chords = Half of the sum of intercepted arcs.

Let $\theta$ be the angle between the intersection of chords AC and BD.

$\theta +112^\circ=180^\circ$ (Linear pair)

$\theta =180^\circ-112^\circ$

$\theta =68^\circ$

Using the angle of intersecting chords theorem, we get

$\theta=\dfrac{arc(AB)+arc(CD)}{2}$

$68^\circ =\dfrac{y^\circ+(80+y)^\circ}{2}$

$136^\circ =(80+2y)^\circ$

On simplification, we get

$136=80+2y$

$136-80=2y$

$56=2y$

Divide both sides by 2.

$28=y$

Now,

$arc(DC)=(80+y)^\circ$

$arc(DC)=(80+28)^\circ$

$arc(DC)=108^\circ$

Therefore, the correct option is C.

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3 years ago

Letter A represents the decimal

Sauron [17]

Answer:

answer is 0.4

Step-by-step explanation:

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3 years ago

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A box with a square base and open top must have a volume of 157216 cm3. We wish to find the dimensions of the box that minimize

shepuryov [24]

Answer:

Base Length of 68cm
Height of 34 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume $=x^2h=157216$

$h=\dfrac{157216}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{157216}{x^2}\\A(x)=x^2+4x\left(\dfrac{157216}{x^2}\right)\\A(x)=\dfrac{x^3+628864}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+628864}{x}\\A'(x)=\dfrac{2x^3-628864}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68$

Step 4: Verify that x=68 is a minimum value

We use the second derivative test

$If\:A(x)=\dfrac{x^3+628864}{x}\\A''(x)=\dfrac{2x^3+1257728}{x^3}\\$When x=68\\A''(x)=6$

Since the second derivative is positive at x=68, then it is a minimum point.

Recall:

$h=\dfrac{157216}{x^2}\\h=\dfrac{157216}{68^2}=34$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 68cm
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3 years ago

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307 + 69x

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342 + 68x

Since they are buying the same number of rolls to cover the same area, then

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It’s going to be C!!

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