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kondaur [170]
3 years ago
9

Which two points satisfy y = -x2 + 2x + 4 and x + y = 4?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
8 0
Y=4-x
0= -x^2 + 3x
x=0 or 3
so y = 4-0 or y= 4-3
y=4 or 1

The two points are:
(0,4) and (3,1)
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What are the real zeros of the function g(x) = x^3 + 2x^2 − x − 2?
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If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial.  C is our constant, 2, and d is the leading coefficient, 1.  The factors of 2 are +/- 1 and +/-2.  The factors for 1 are +/-1 only.  Meaning, in all, there are 4 possibilities as roots for this polynomial.  But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above.  Let's try x = -1, factor form (x + 1).  If there is no remainder when we do the synthetic division, then -1 is a root.  Put -1 outside the "box" and the coefficients from the polynomial inside: -1  (1  2  -1  -2).  Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1.  Put that -1 up under the 2 and add to get 1.  Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2.  -1 times -2 is 2, and -2 + 2 = 0.  So we have our first root of (x+1).  The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!).  The new polynomial is x^{2} +x-2=0.  That can also be factored to find the remaining 2 roots.  Use standard factoring to find that the other 2 solutions are (x+2) and (x-1).  Our solutions then are x = -2, -1, 1, choice B from above.
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3 years ago
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a hiker climbs down a 90 foot hill in five minutes what is his average change in elevation per minute
andrey2020 [161]

90 feet/5 minutes = 18 feet per minute

3 0
3 years ago
The upper arm length of females over 20 years old in a country is approximately Normal with mean 35.8 centimeters (cm) and stand
lana [24]

Answer:

a) The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

b) 16% of women over 20 have upper arm lengths less than 33.3 cm.

Step-by-step explanation:

The Empirical Rule(68-95-99.7 Rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 35.8 cm

Standard deviation = 2.5 cm

(a) What range of lengths covers almost all (99.7%) of this distribution?

This range is from 3 standard deviations below the mean to three standard deviations above the mean.

So from 35.8 - 3*2.5 = 28.3 cm to 35.8 + 3*2.5 = 43.3 cm

The range of lengths from 28.3 cm to 43.3 cm covers almost all (99.7%) of this distribution.

(b) What percent of women over 20 have upper arm lengths less than 33.3 cm?

68% of the women over 20 have upper arm length between 33.3 cm and 38.3 cm. The other 32% have upper arm length lower than 33.3 cm or higher than 38.3. The distribution is symmetric, so 16% of the have upper arm length lower than 33.3 cm and 16% have upper arm length higher than 38.3 cm

So 16% of women over 20 have upper arm lengths less than 33.3 cm.

4 0
3 years ago
A student earned grades of 79, 74, and 73 on her three regular exams. She earned a grade of 66 on the final exam and 88 on her c
navik [9.2K]

Answer:

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Step-by-step explanation:

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Weighted mean grade

w = \dfrac{\sum wx}{\sum w}

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