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2 years ago

Which description is correct for this polynomial 4x^2+3x+5?

Mathematics

1 answer:

mash [69]2 years ago

7 0

Its a quadratic trinomial

Its quadratic because the highest degree in the equation is two

Its a trinomial because there are three different terms in the equation

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Svet_ta [14]

Answer:

Step-by-step explanation:

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2 years ago

Middle school Math!!!

makvit [3.9K]

Y=1x+2 (you don’t have to write the 1 in front of the x so you can write it as y=x+1)

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2 years ago

1. 1*1/6=___

Ratling [72]

Step-by-step explanation:

1. 1*1/6= 1 /6=1.67

2. 9*7/10= 63/10= 6.3

3. 7*4/8= 28/8= 3.5

4. 1/2 of 2= 1/2*2= 0.5*2= 1

5. 1/12 of 2= 1/12*2= 8.33*2= 16.66

6.2/6 of 2= 2/6*2= 3.33*2= 6.66

7. 1/3 of 5= 1/3*5= 3.33*5= 16.65

8. 3/10 of 8= 3/10*8= 0.3*8= 2.4

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2 years ago

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Anastaziya [24]

A y-intercept always has to have an x-value of 0 so the answer is D (the last option)

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2 years ago

Read 2 more answers

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

2 years ago