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4 years ago

WORTH 15 POINTS NEED NOW Factor: (x – 4)^2 = 25

1 answer:

3 0

The answer is
$(x - 4) ^{2} \: = 25 \\ {x}^{2} - 8x + {4}^{2} - 25 = 0 \\ {x}^{2} - 8x - 9 = 0 \\ x = 9 \: or \: x = - 1$

Hope it helps!

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The rabbit population in a certain area is 500% of last year's population. There are 1,100 rabbits this year. How many were th

andreev551 [17]

Answer:

5,500 rabbits

Step-by-step explanation:

500% of something is 5 times the original amount. 1,100 rabbits times 5 equals 5,500.

8 0

4 years ago

What does 100%, 4/4 ,1.000 have in common

Vadim26 [7]

They all end up being 1
100% = 1
4/4 = 1
1.000 = 1

4 0

3 years ago

Read 2 more answers

In a sample of students, 12 out of 24 students in a class wanted a sandwich for lunch. Using the data from the sample, predict h

natka813 [3]

Answer:

200

Step-by-step explanation:

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8 0

3 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

3 years ago

Example 2:

Montano1993 [528]

Taking this example into account, we can see that setting the first value equal to 1, we obtain that $F(x)=0.5x+1=4$ and $x=6$ . Using this information, we find that $F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5$ . It shows that when x is positive, the succussive terms are increasing.

Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$ . In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$ . By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.

Using the function $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$ . If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$ . In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that $g(x)=-x+2=1$ and x=1. In the next iteration, $g(x+1)=-(x+1)+2=0$ and the iterations are decreasing.

8 0

4 years ago