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Katen [24]
2 years ago
11

If one of the roots of the equation x²- 4x + k=0exceeds the other by 2, then find the roots and determine the value of k

Mathematics
1 answer:
BaLLatris [955]2 years ago
6 0

Answer:

roots: 1 and 3

k = 3

Step-by-step explanation:

2 roots: p and p+2

(x-p) (x-p-2) = x² - 4x + k

x² -2px -2x + p² + 2p = x² - 2 (p+1)x + (p² + 2p) = x² -4x + k

-2 (p+1) = -4

p+1 = 2

p = 1   ... root 1

p' = 1+2 = 3   ... root 2

k = p² + 2p = 3

check: (x-1) (x-3) = x² - 4x + 3 = x² - 4x + k

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fredd [130]
<h3>Answer: Choice B</h3>

Angle 1 = 147 degrees

Angle 2 = 80 degrees

Angle 3 = 148 degrees

======================================================

Work Shown:

(angle 1) + 33 = 180

angle 1 = 180-33

angle 1 = 147 degrees

---------------

Focus on the left most triangle that has angles 33 and 47 as interior angles. The missing angle is 180-33-47 = 100 degrees

The angle exterior to this 100 degree angle is angle 2

angle 2 = 180-100 = 80

We have enough info to conclude the answer must be choice B.

---------------

Let's keep going to find angle 3

The vertical angle for the 100 degree angle is also 100 degrees. This second 100 degree angle is part of the triangle on the right

This triangle on the right has interior angles 100 and 48

The missing interior angle is 180-100-48 = 32

The angle supplementary to this is 180-32 = 148, which is angle 3.

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3 years ago
Help urgent plz!❤️❤️❤️Please please urgently
sleet_krkn [62]

Answer:

Problem 2): \left \{x\, |\,0\leq x\leq 240\,\,\right \}

which agrees with answer C listed.

Problem 3) :  D = (-3, 6]  and R = [-5, 7]

which agrees with answer D listed

Step-by-step explanation:

Problem 2)

The Domain is the set of real numbers in which the function (given by a graph in this case) is defined. We see from the graph that the line is defined for all x values between 0 and 240. Such set, expressed in "set builder notation" is:

\left \{x\, |\,0\leq x\leq 240\,\right \}

Problem 3)

notice that the function contains information on the end points to specify which end-point should be included and which one should not. The one on the left (for x = -3 is an open dot, indicating that it should not be included in the function's definition, therefor the Domain starts at values of x strictly larger than -3. So we use the "parenthesis" delimiter in the interval notation for this end-point. On the other hand, the end point on the right is a solid dot, indicating that it should be included in the function's definition, then we use the "square bracket notation for that end-point when writing the Domain set in interval notation:

Domain = (-3, 6]

For the Range (the set of all those y-values connected to points in the Domain) we use the interval notation form:

Range = [-5, 7]

since there minimum y-value observed for the function is at -5 , and the maximum is at 7, with a continuum in between.

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2 years ago
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Answer:

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3 years ago
X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
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