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3 years ago

A triangle has two side lengths 7 and 9. What value could the length of the third side be? Check all that apply

Mathematics

2 answers:

Licemer1 [7]3 years ago

7 0

Answer:

$8,5,10,13$

Step-by-step explanation:

Let

x-----> the length of the third side

we know that

The Triangle Inequality Theorem states that the sum of any two sides of a triangle must be greater than the measure of the third side

case A)

$7+9 >x\\16> x\\x$

case B)

$7+x >9\\x>9-7\\ x>2\ units$

The solution for the length of the third side is the interval------> $(2,16)$

All real numbers greater than $2$ and less than $16$

therefore

the solutions are

$8,5,10,13$

mamaluj [8]3 years ago

3 0

Each side has to be less then the sum of the other two sides.
A, B, C, and E work.

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Answer:

x-intercept: (-4,0)

y-intercept: (0,-2)

Step-by-step explanation:

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2 years ago

What are the roots of the quadratic function in the graph?

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3 years ago

The width of a container is 5 feet less than its height. Its length is 1 foot longer than its height. The volume of the containe

juin [17]

The height of container is 8 feet

Solution:

Let "w" be the width of container

Let "l" be the length of container

Let "h" be the height of container

The width of a container is 5 feet less than its height

Therefore,

width = height - 5

w = h - 5 ------ eqn 1

Its length is 1 foot longer than its height

length = 1 + height

l = 1 + h ---------- eqn 2

The volume of container is given as:

$v = length \times width \times height$

Given that volume of the container is 216 cubic feet

$216 = l \times w \times h$

Substitute eqn 1 and eqn 2 in above formula

$216 = (1 + h) \times (h-5) \times h\\\\216 = (h+h^2)(h-5)\\\\216 = h^2-5h+h^3-5h^2\\\\216 = h^3-4h^2-5h\\\\h^3-4h^2-5h-216 = 0$

Solve by factoring

$(h-8)(h^2+4h+27) = 0$

Use the zero factor principle

If ab = 0 then a = 0 or b = 0 ( or both a = 0 and b = 0)

Therefore,

$h - 8 = 0\\\\h = 8$

Also,

$h^2+4h+27 = 0$

Solve by quadratic equation formula

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\mathrm{For\:} a=1,\:b=4,\:c=27:\quad h=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:27}}{2\cdot \:1}$

$h = \frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:27}}{2}=\frac{-4+\sqrt{92}i}{2}$

Therefore, on solving we get,

$h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i$

Thus solutions of "h" are:

h = 8

$h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i$

"h" cannot be a imaginary value

Thus the solution is h = 8

Thus the height of container is 8 feet

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3 years ago

To the nearest degree what is the measure of each exterior angle of a regular hexagon ?

otez555 [7]

This is the concept of geometry, the total sum of angles in a hexagon is given by:
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N/B
the exterior and interior angles are supplementary to each other.
Therefore the size of the exterior angle will be:
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the answer is A

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