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noname [10]
2 years ago
14

Solve the following quadratic function by utilizing the square root method. f(x) = 1 - x2 x = + [?]

Mathematics
2 answers:
faltersainse [42]2 years ago
7 0

Answer:

Step 1) y=16-x2. Swap the sides so that all terms of the variables are on the left side. Step 2) 16-x_{2}=y. Subtract 16 from both sides. Step 3) -x_{2}=y-16 Divide the two sides by -1. Step 4). \frac{-x_{2}}{-1}=\frac{y-16}{-1} Dividing by -1 undoes the multiplication by -1. Step 5). x_{2}=\frac{y-16}{-1} Step 6) dived y-16 by -1 And the final answer = x_{2}=16-y

Step-by-step explanation:

Triss [41]2 years ago
4 0

Answer: 1

Step-by-step explanation:

If x = 0

then, f(x) = 1 - 0

f(x) = 1

x = ±1

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Given the function f(x)=4x^8+7x^7+1x^6+1 What is the value of f(−2)?
qwelly [4]
<span> f(x)=4x^8+7x^7+1x^6+
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7 0
3 years ago
The quotient of 6 times a number and 16
alexandr1967 [171]
6x=16
÷6 both sides
x=16/6 or 8/3
4 0
3 years ago
What is the 4th equivalent fraction to <br> 5/6
Lady bird [3.3K]

Answer:

20/24

Step-by-step explanation: Hope this helps

3 0
2 years ago
Discuss the validity of the following statement. If the statement is always​ true, explain why. If​ not, give a counterexample.
Zarrin [17]

Correction:

Because F is not present in the statement, instead of working on​P(E)P(F) = P(E∩F), I worked on

P(E∩E') = P(E)P(E').

Answer:

The case is not always true.

Step-by-step explanation:

Given that the odds for E equals the odds against E', then it is correct to say that the E and E' do not intersect.

And for any two mutually exclusive events, E and E',

P(E∩E') = 0

Suppose P(E) is not equal to zero, and P(E') is not equal to zero, then

P(E)P(E') cannot be equal to zero.

So

P(E)P(E') ≠ 0

This makes P(E∩E') different from P(E)P(E')

Therefore,

P(E∩E') ≠ P(E)P(E') in this case.

8 0
3 years ago
Write the standard form of the equation of the line through the given point with the given slope. Through (-2,5), slope =-4
AleksandrR [38]

Answer:

\huge\boxed{4x+y=-3}

Step-by-step explanation:

The point-slope form of an equation of a line:

y-y_1=m(x-x_1)

<em>(x₁, y₁)</em><em> - point on a line</em>

<em>m</em><em> - slope</em>

<em />

We have

m=-4,\ (-2,\ 5)\to x_1=-2,\ y_1=5

Substitute:

y-5=-4(x-(-2))\\\\y-5=-4(x+2)

Convert to the standard form

Ax+By=C

y-5=-4(x+2)                <em> use the distributive property</em>

y-5=(-4)(x)+(-4)(2)

y-5=-4x-8               <em>add 5 to both sides</em>

y-5+5=-4x-8+5

y=-4x-3               <em>add 4x to both sides</em>

4x+y=-4x+4x-3\\\\4x+y=-3

3 0
3 years ago
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