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3 years ago

An iscoclees triangle may be right or acute but never obtuse

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

4 0

Answer:

An isosceles triangle is a triangle in which the two sides of the polygon are equal as well as the angles opposite to these sides are also corresponding

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I need the answer fast

Fantom [35]

Answer:

B. What was the highest temperature this month?

Step-by-step explanation:

Rory is recording the highest temperature for the duration of the day for month's worth of data.

B. is the easiest one to answer, as you simply will have to look for the greatest number within the set, and that will be the answer.

A. is too broad, and does not imply that only the "high temperature" is needed, rather just a leveled temperature within day in that month.

C. is the opposite of what Rory had recorded, and without data, he cannot answer it.

D. is answerable as well, but, again, the only temperature recorded is the high temperature. Temperature can fluctuate depending on the time of the day, and can be in the 50s one hour and the 90s in another.

5 0

2 years ago

Write an equation for a quadratic function that has x intercepts (-3, 0) and (5, 0)

Blizzard [7]

Answer:

One possible equation is $f(x) = (x + 3)\, (x - 5)$ , which is equivalent to $f(x) = x^{2} - 2\, x - 15$ .

Step-by-step explanation:

The factor theorem states that if $x = x_{0}$ (where $x_{0}$ is a constant) is a root of a function, $(x - x_{0})$ would be a factor of that function.

The question states that $(-3,\, 0)$ and $(5,\, 0)$ are $x$ -intercepts of this function. In other words, $x = -3$ and $x = 5$ would both set the value of this quadratic function to $0$ . Thus, $x = -3\!$ and $x = 5\!$ would be two roots of this function.

By the factor theorem, $(x - (-3))$ and $(x - 5)$ would be two factors of this function.

Because the function in this question is quadratic, $(x - (-3))$ and $(x - 5)$ would be the only two factors of this function. In other words, for some constant $a$ ( $a \ne 0$ ):

$f(x) = a\, (x - (-3))\, (x - 5)$ .

Simplify to obtain:

$f(x) = a\, (x + 3)\, (x - 5)$ .

Expand this expression to obtain:

$f(x) = a\, (x^{2} - 2\, x - 15)$ .

(Quadratic functions are polynomials of degree two. If this function has any factor other than $(x - (-3))$ and $(x - 5)$ , expanding the expression would give a polynomial of degree at least three- not quadratic.)

Every non-zero value of $a$ corresponds to a distinct quadratic function with $x$ -intercepts $(-3,\, 0)$ and $(5,\, 0)$ . For example, with $a = 1$ :