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8090 [49]
3 years ago
10

2x to the fourth power + 5x squared + 3, factored

Mathematics
2 answers:
Murljashka [212]3 years ago
8 0
We want to factor:
2x^4+5x^2+3

This may seem tricky because of the high powers, but remember that x^4=(x^2)^2. Then we can factor it as:
2x^4+5x^2+3=(2x^2+3)(x^2+1)
Alexeev081 [22]3 years ago
8 0
(x squared+1)(2x squared+3)
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Please help me with this please is hard
Elan Coil [88]

Answer:

1. $5

2. $38.4

Step-by-step explanation:

1) $5

2) 120/100 x 32 = $38.4

4 0
3 years ago
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The distance possible to sea on a clear day varies directly as the square roo of the height of six metres above sea level, it is
Natali [406]

Answer:

90 km is the answer of your question

4 0
3 years ago
Find the y-intercept of the following equation. Simplify your answer. 4x + y = 7
Bogdan [553]
4x + y = 7
y = -4x + 7

using y = mx + b

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5 0
3 years ago
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Approximate the solution to the equation above using three iterations of successive approximation. Use the graph below as a star
n200080 [17]

Answer:

<em>x=25/16</em>

Step-by-step explanation:

<u>Approximate Roots of Functions:</u>

We are required to find the root of  

\displaystyle 5^{-x}+7=\ 2x+4

Let's construct a function

\displaystyle f(x)=\ 5^{-x}+7-2x-4

Reducing:

\displaystyle f(x)= 5^{-x}-2x+3

The equation will have a solution when f(x)=0. We'll start from the approximate crossing point given in the graph (x=1.5)

\displaystyle f(1.5)=5^{-1.5}-2(1.5)+3\ =0.089

We'll move up on x=1.52  

\displaystyle f(1.52)=5^{-1.52}-2(1.52)+3=\ 0.046

We are closer to the solution, let's try some more up

\displaystyle f(1.54)=5^{-1.54}-2(1.54)+3=\ 0.0038

This value x=1.54 is close enough to the solution of the original equation

From the options given, \frac{25}{16} is the closest to our solution

\boxed{Answer:\ x=25/16}

8 0
4 years ago
The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered. The delivery tim
Hatshy [7]

Answer:

The expected cost is 152

Step-by-step explanation:

Recall that since Y is uniformly distributed over the interval [1,5] we have the following probability density function for Y

f_Y(y ) = \frac{1}{5-1} = \frac{1}{4} if 1\leq y \leq 5 and 0 othewise. (To check this is the pdf, check the definition of an uniform random variable)

Recall that, by definition  

E(Y^k) = \int_{-\infty}^{\infty} y^kf_Y(y)dy

Also, we are given that C = 50+3Y+9Y^2. Recall the following properties of the expected value. If X,Y are random variables, then

E(a+bX+cY) = a+bE(X)+cE(Y)

Then, using this property we have that E[C] = 50+3E[Y]+ 9E[Y^2].

Thus, we must calculate E[Y] and E[Y^2].

Using the definition, we get that

E[Y] = \int_{1}^{5}\frac{y}{4} dy =\frac{1}{4}\left\frac{y^2}{2}\right|_{1}^{5} = \frac{25}{8}-\frac{1}{8} = 3

E[Y^2] = \int_{1}^{5}\frac{y^2}{4} dy =\frac{1}{4}\left\frac{y^3}{3}\right|_{1}^{5} = \frac{125}{12}-\frac{1}{12} = \frac{31}{3}

Then

E(C) = 50 + 3\cdot 3 + 9 \cdot \frac{31}{3} = 152

5 0
3 years ago
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