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Cloud [144]
3 years ago
10

Consider the absolute value inequality: |x - 6| > 20. Will the graph of the solutions to the inequality result in an "and" or

an "or" inequality? Without solving to find the exact solutions, explain
Mathematics
1 answer:
Dmitrij [34]3 years ago
3 0

Step-by-step explanation:

Case 1 : |x| > a, => x > a or x < -a

Case 2 : |x| < a, => -a < x < a

Since this question follows Case 1, we will have an "or" inequality.

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In the expression (ax+b)*(cx^2+dx+e), find two set of values of a,b,c,d and e so that the coefficient of x^3and x terms after ex
Brilliant_brown [7]

Answer:

2, 0, 2, 3, 5

1, 2, 4, 0, 5

Step-by-step explanation:

(ax + b)(cx² + dx + e)

acx³ + adx² + aex + bcx² + bdx + be

2(2)x³ + 2(3)x² + 2(5)x + 0 + 0 + 0

4x³ + 6x² + 10x

a = 2

b = 0

c = 2

d = 3

e = 5

1(4)x³ + 1(0)x² + 1(10)x + 2(4)x² + 0 + 10

4x³ + 8x² + 10x + 10

a = 1

b = 2

c = 4

d = 0

e = 5

7 0
2 years ago
Which of the following is true?
gtnhenbr [62]

Answer:

C..... I think it is a answer

6 0
3 years ago
Which equation represents a hyperbola shown in the graph?
Paladinen [302]

Answer:

It's C

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
What fraction would you add to 29/6 to make 4
Zigmanuir [339]
Keeping in mind that 29/6 is greater than 4, is actually 4 and 5/6, so the amount we'll "add" will be a negative one.

\bf \cfrac{29}{6}+x=4\implies x=4-\cfrac{29}{6}\implies x=\cfrac{4}{1}-\cfrac{29}{6}\implies x=\cfrac{(6)4-(1)29}{6}&#10;\\\\\\&#10;x=\cfrac{24-29}{6}\implies x=\cfrac{-5}{6}\\\\&#10;-------------------------------\\\\&#10;\cfrac{29}{6}+\left( -\cfrac{5}{6} \right)=4
8 0
4 years ago
if 18 meters of high resistance wire with a diameter of 1.2 mm has a resistance of 10 ohms, what is the resistance of 27 m of th
Stells [14]
The electrical resistance of a wire varies as its length and inversely as the square of the diameter.
R = (k*L)/(d^2)
where k = proportionality constant
Since the two wires have the same material, their proportionality constant is same.
Equating that
(R1*d1^2)/L1 = (R2*d2^2)/L2

Given that R1 = 10 ohms, d1 = 1.2 mm or 0.0012 m, L1 = 18 m, d2 = 1.5 mm or 0.0015 m, L2 = 27 m, and R2 is unknown.
Therefore
[10*(0.0012^2)]/18 = [R2*(0.0015^2)]/27
R2 = 9.6 ohms 
6 0
3 years ago
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