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3 years ago

The three side lengths of a triangle are given. Which triangle is a right triangle?

Mathematics

1 answer:

EastWind [94]3 years ago

7 0

Answer:

Triangle A

Explanation:

This isn't really an explanation, but I don't really know how to explain. The other answers just aren't correct (They aren't right triangles).

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The equation c=2.5t represents the cost,(c),for (t) ticket to the school play. Does a value of 3.5 for (t) make sense in this si

scoundrel [369]

Answer:

Step-by-step explanation:

No, this doesn't make sense because as stated in the problem, each ticket costs $2.50. Since you cant buy part of a ticket, you would have to have a whole number for t. You cannot buy 3.5 tickets.

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3 years ago

For each equation how many solutions why? Question e

Kay [80]

Answer:

one real solution

Step-by-step explanation:

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3 years ago

It snowed for 11 days in January and 9 days in February. How many days did it snow in all?

m_a_m_a [10]

Answer:

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Step-by-step explanation:

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3 years ago

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I can’t figure out how to use the zeros in the polynomial. Please explain

Temka [501]

Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

---------------------------------------

$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

--------------------------------------

$x = 6\\\\(x-6) = 0$

--------------------------------------

$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

--------------------------------------

$x = 1\\\\(x-1) = 0$

--------------------------------------

$x = 4\\\\(x-4) = 0$

----------------------------------------

$x = -1\\\\(x + 1) = 0$

-----------------------------------------

$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

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3 years ago