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Lelechka [254]
3 years ago
10

Given f(x) 2^x -7 find the x intercepts

Mathematics
1 answer:
brilliants [131]3 years ago
7 0
Be sure to include the "=" sign:   <span>f(x) = 2^x -7 

To find the x-intercepts, set f(x) = 0 and solve for x:  2^x - 7 = 0,
or
2^x = 7

Take the common log of both sides:  x log 2 = log 7
                              log 7
Solve for x:     x = --------- = 2.81 (approx)
                               log 2

</span>
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The 2nd question (the answer is c) but I don't know why so can someone solve this for me ?
Genrish500 [490]
The formula for surface area of a sphere is A=4*PI*r^2

Using that formula and the known surface are we can find the radius. We can then use that to find the volume.

500 = 4*PI*r^2

Divide both sides by 4:
500/4 = PI *r^2
125 = PI *r^2

Divide both sides by PI

125/PI = r^2
39.79 = r^2
r = SQRT(39.79)
r = 6.3 cm

Now we can find volume using V=4/3*PI*r^3

V = 4/3 *PI *6.3^3

V= 1051.3

The answer is C
7 0
3 years ago
A student claims that 2i is the only imaginary root of a polynomial equation that has real coefficients. Explain the student's m
____ [38]

Answer:

The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i

Step-by-step explanation:

1) This claim is mistaken.

2) The Fundamental Theorem of Algebra assures that any polynomial  f(x)=0 whose degree is n ≥1 has at least one Real or Imaginary root. So by the Theorem we have infinitely solutions, including imaginary roots ≠ 2i with real coefficients.

a_{0}x^{n}+a_{1}x^{2}+....a_{1}x+a_{0}

For example:

3) Every time a polynomial equation, like a quadratic equation which is an univariate polynomial one, has its discriminant following this rule:

\Delta < 0\\b^{2}-4*a*c

We'll have <em>n </em>different complex roots, not necessarily 2i.

For example:

Taking 3 polynomial equations with real coefficients, with

\Delta < 0

-4x^2-x-2=0 \Rightarrow S=\left \{ x'=-\frac{1}{8}-i\frac{\sqrt{31}}{8},\:x''=-\frac{1}{8}+i\frac{\sqrt{31}}{8} \right \}\\-x^2-x-8=0 \Rightarrow S=\left\{\quad x'=-\frac{1}{2}-i\frac{\sqrt{31}}{2},\:x''=-\frac{1}{2}+i\frac{\sqrt{31}}{2} \right \}\\x^2-x+30=0\Rightarrow S=\left \{ x'=\frac{1}{2}+i\frac{\sqrt{119}}{2},\:x''=\frac{1}{2}-i\frac{\sqrt{119}}{2} \right \}\\(...)

2.2) For other Polynomial equations with real coefficients we can see other complex roots ≠ 2i. In this one we have also -2i

x^5\:-\:x^4\:+\:x^3\:-\:x^2\:-\:12x\:+\:12=0 \Rightarrow S=\left \{ x_{1}=1,\:x_{2}=-\sqrt{3},\:x_{3}=\sqrt{3},\:x_{4}=2i,\:x_{5}=-2i \right \}\\

4 0
3 years ago
vanesa multiplies two integers. When she adds the product to 4, the sum is 0. What can you say about the two integers
Lelu [443]
The Integers Are Positive They Are Not Negative Numbers. If It Was Negative It Would Be -2,-3,-4. While If It's Positive It Should Be 0,1,2,3,4 And So On.
4 0
3 years ago
Simplify the radical expression.<br><br> square root of 128
neonofarm [45]
The square root of 128 is 11.3
4 0
3 years ago
Which numbers are roots of the polynomial function shown? Check all that apply.
posledela

Answer:

-4

2

6

Are the answers to this question. i just did this on edge..


4 0
3 years ago
Read 2 more answers
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