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lapo4ka [179]
3 years ago
12

Triangle ABC has 2 known sides. A = 7, b = 5. Angle C = 24.63. What is the value of side c, correct to one decimal place?

Mathematics
1 answer:
Elis [28]3 years ago
5 0
We use the Law of Cosines:
side c² = a² + b² -2ab • cos(C)
side c^2 = 7^2 + 5^2 -(2*7*5*cosine(24.63))
side c^2 = 49 + 25 -(70 * 0.90902)
side c^2 = 74 - <span> <span> <span> 63.6314 </span> </span> </span>
side c^2 = <span> <span> <span> 10.3686 </span> </span> </span>
side c = <span> <span> <span> 3.22

Source:
http://www.1728.org/trigtut2.htm

</span></span></span>
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7 0
3 years ago
If y is directly proportional to x and when y = 35, x = 140, what is the value of x when y = 70?
levacccp [35]

Answer: 280 = x


Step-by-step explanation: Okay you are using this equation! y = kx But you must rearrange it!

When rearranged you get y/x = k

Plug in x and y!

y/x = k

35/140 = k

0.25 = k

Now find the value of x.

You use this formula now, y/k = x

Plug in k from the last problem and y = 70

y/k = x

70/0.25 = x

280 = x



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4 years ago
Solve for X
sergij07 [2.7K]

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Step-by-step explanation:

8 0
3 years ago
You are creating an open top box with a piece of cardboard that is 16 x 30“. What size of square should be cut out of each corne
Arada [10]

Answer:

\frac{10}{3} \ inches of square should be cut out of each corner to create a box with the largest volume.

Step-by-step explanation:

Given: Dimension of cardboard= 16 x 30“.

As per the dimension given, we know Lenght is 30 inches and width is 16 inches. Also the cardboard has 4 corners which should be cut out.

Lets assume the cut out size of each corner be "x".

∴ Size of cardboard after 4 corner will be cut out is:

Length (l)= 30-2x

Width (w)= 16-2x

Height (h)= x

Now, finding the volume of box after 4 corner been cut out.

Formula; Volume (v)= l\times w\times h

Volume(v)= (30-2x)\times (16-2x)\times x

Using distributive property of multiplication

⇒ Volume(v)= 4x^{3} -92x^{2} +480x

Next using differentiative method to find box largest volume, we will have \frac{dv}{dx}= 0

\frac{d (4x^{3} -92x^{2} +480x)}{dx} = \frac{dv}{dx}

Differentiating the value

⇒\frac{dv}{dx} = 12x^{2} -184x+480

taking out 12 as common in the equation and subtituting the value.

⇒ 0= 12(x^{2} -\frac{46x}{3} +40)

solving quadratic equation inside the parenthesis.

⇒12(x^{2} -12x-\frac{10x}{x} +40)=0

Dividing 12 on both side

⇒[x(x-12)-\frac{10}{3} (x-12)]= 0

We can again take common as (x-12).

⇒ x(x-12)[x-\frac{10}{3} ]=0

∴(x-\frac{10}{3} ) (x-12)= 0

We have two value for x, which is 12 and \frac{10}{3}

12 is invalid as, w= (16-2x)= 16-2\times 12

∴ 24 inches can not be cut out of 16 inches width.

Hence, the cut out size from cardboard is \frac{10}{3}\ inches

Now, subtituting the value of x to find volume of the box.

Volume(v)= (30-2x)\times (16-2x)\times x

⇒ Volume(v)= (30-2\times \frac{10}{3} )\times (16-2\times \frac{10}{3})\times \frac{10}{3}

⇒ Volume(v)= (30-\frac{20}{3} ) (16-\frac{20}{3}) (\frac{10}{3} )

∴  Volume(v)= 725.93 inches³

6 0
3 years ago
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