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3 years ago

Identify as a monomial, binomial, or trinomial. 2x^2+5x-8

Mathematics

2 answers:

swat323 years ago

6 0

Answer:

<h2>TRINOMIAL</h2>

Step-by-step explanation:

2x² + 5x - 8 there are no like terms

Three terms: 2x², 5x and (-8).

Therefore it's a trinomial.

Monomial:

x, 3, 3y, -9y² - one term

Binomial:

2 + a. xy + y³, 3x²y² - xy - two terms

Korvikt [17]3 years ago

3 0

Answer:

trinomial

Step-by-step explanation:

The expression is a trinomial because it consists of three terms (that's the definition of a trinomial. The three terms are: 2x^2, 5x, and -8.

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Six years ago annabelle was twice as old as jason was. now, annabelle is four years older than jason. how old is jason now?

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3 years ago

Which equation represents a circle with a center at (-4,9) and a diameter of 10 units?

Dennis_Churaev [7]

Answer:

$(x+4)^2 + (y-9)^2 = 25$

Step-by-step explanation:

To write the equation of a circle, use the formula $(x-h)^2+(y-k)^2 = r^2$ where the center of the circle is (h, k).

This means the equation is $(x--4)^2 + (y-9)^2 = r^2\\\(x+4)^2 + (y-9)^2 = r^2$ .

Find the radius r by dividing the diameter of 10 in 2. The radius is 5.

So the equation of the circle is $(x+4)^2 + (y-9)^2 = 25$

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3 years ago

Let Y1 and Y2 be independent exponentially distributed random variables, each with mean 7. Find P(Y1 > Y2 | Y1 < 2Y2). (En

ArbitrLikvidat [17]

Y₁ and Y₂ are independent, so their joint density is

$f_{Y_1,Y_2}(y_1,y_2)=f_{Y_1}(y_1)f_{Y_2}(y_2)=\begin{cases}\frac1{49}e^{-\frac{y_1+y_2}7}&\text{for }y_1\ge0,y_2\ge0\\0&\text{otherwise}\end{cases}$

By definition of conditional probability,

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = P((Y₁ > Y₂) and (Y₁ < 2 Y₂)) / P(Y₁ < 2 Y₂)

Use the joint density to compute the component probabilities:

• numerator:

$P((Y_1>Y_2)\text{ and }(Y_1$

$=\displaystyle\frac1{49}\int_0^\infty\int_{\frac{y_1}2}^{y_1}e^{-\frac{y_1+y_2}7}\,\mathrm dy_2\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\int_{-\frac{3y_1}{14}}^{-\frac{2y_1}7}e^u\,\mathrm du\,\mathrm dy_1$

$=\displaystyle-\frac17\int_0^\infty\left(e^{-\frac{2y_1}7} - e^{-\frac{3y_1}{14}}\right)\,\mathrm dy_1$

$=\displaystyle-\frac17\left(-\frac72e^{-\frac{2y_1}7} + \frac{14}3 e^{-\frac{3y_1}{14}}\right)\bigg|_0^\infty$

$=\displaystyle-\frac17\left(\frac72 - \frac{14}3\right)=\frac16$

• denominator:

$P(Y_1$

(I leave the details of the second integral to you)

Then you should end up with

P(Y₁ > Y₂ | Y₁ < 2 Y₂) = (1/6) / (2/3) = 1/4

5 0

2 years ago