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3 years ago

14

Suppose a triangle has two sides of length 33 and 37 and that the angle between these two sides is 120 what is the length of the

third side of the triangle

1 answer:

Nutka1998 [239]3 years ago

5 0

Answer:

60.65

Step-by-step explanation:

The Law of Cosines can help you figure this out. Call the given sides "a" and "b" and the given angle "C". Then the third side, "c" will satisfy the relation ...

c² = a² + b² -2ab·cos(C)

= 33² +37² -2·33·37·cos(120°) = 3679

c = √3679 ≈ 60.65476 ≈ 60.65

The length of the third side is about 60.65 units.

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If a solid has faces that consist of 2 equilateral triangles and 3 congruent rectangles, what type of solid is it?

svlad2 [7]

A right triangular prism is a solid that has faces that consist of 2 equilateral triangles and 3 congruent rectangles. It is also described to have two parallel faces and 3 other faces that belong to the same place (and are not parallel to the parallel pair of faces).

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3 years ago

Abby has a jewelry business. She uses small square boxes to ship her homemade rings to stores to sell. The ring box measures 12

aleksandr82 [10.1K]

Answer:

Volume of Shipping box= lenghth × breadth × height

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Step-by-step explanation:

Volume of ring box = 12×12×12

=1728 units

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Hope am correct

Thanks

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3 years ago

Use special right triangle find the exact value of x. <br> X=?

swat32

Answer: 12

Step-by-step explanation:

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3 years ago

If two straight lines intersect and one angle is 30 degrees what are the other three angles

mel-nik [20]

Easey peasy
oposite angle is equal
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oposite angle=30
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30+other=180
subtract 30
other=150
other is oposite other other
other=other other
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angles are
150,30,150

3 0

2 years ago

Read 2 more answers

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

<u>For the first parenthesis:</u>

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

<u>For the second parenthesis:</u>

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

8 0

3 years ago