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castortr0y [4]
3 years ago
9

Is ( 1,-1) a solution to the following system of inequalities:

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

No

Step-by-step explanation:

x-y less than -2

4x+y less than 3

4x less than 3+y

4x less than 3y

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The math club is selling gift wrap for a fumdraiser. They sold all 45 rolls of solid wrapping paper at $4 each and rolls of patt
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17 rolls

Step-by-step explanation:

4 x 45 = 180  

265-180 = 85      

85 ÷ 5 = 17  

hope this helps ;)

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4 years ago
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When number is multiplied by 4 subtracting from 68, the result obtained is the same as three times the sum of the number and 4 f
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Answer:

Let the no.be X

when multiplied by 4 = 4X

68-4X=3(X+4)

68-4X=3X+12

collecting like terms together,

7X=56

therefore; X=8

hence ,the number is 8.

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On a bus, there are 5 men, 8 women, 15 boys and 12 girls. Writing in the simplest form, what is the ratio of:
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Answer:

A.5:8

B.8:5

C.5:4

D. 2:3

E.1:8

F.1:1

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3 years ago
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7) Ted has $6.80 in quarters and dimes. The
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g

Step-by-step explanation:

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3 years ago
Determine if the function f(x)=44−x2‾‾‾‾‾‾√ satisfies the Mean Value Theorem on [0, 2]. If so, find all numbers c on the interva
lina2011 [118]

Answer:

The function f(x) satisfies the Mean Value Theorem

Step-by-step explanation:

Mean Value Theorem states that if f be a function such that,

  •  It is continuous on [a, b].
  • It is differentiable on (a, b).

Then there is at least a number c in (a, b) such that,

f'(c) = \frac{f(b) - f(a)}{b-a}

Here, the given function,

f(x) = \sqrt{44-x^2}

∵ f(x) is defined for all real values x for which 44-x² ≥ 0

44 ≥ x² ⇒ ±√44 ≥ x ⇒-√44 ≤ x ≤ √44

Thus, f(x) is continuous on [0, 2],

f'(x) = \frac{1}{2\sqrt{44-x^2}}(-2x) =-\frac{x}{\sqrt{44-x^2}}

∵ f'(x) is defined for all values on the interval (0, 2),

Thus, f(x) is differentiable on (0, 2),

Now,

f'(c) = -\frac{c}{\sqrt{44-c^2}}

\frac{f(2) - f(0)}{2-0}=\frac{\sqrt{44-4}-\sqrt{44}}{2}=\frac{\sqrt{40}-\sqrt{44}}{2}=\sqrt{10}-\sqrt{11}

-\frac{c}{\sqrt{44-c^2}}=\sqrt{10}-\sqrt{11}

\frac{c^2}{44-c^2}=10 + 11 - 2\sqrt{110}

c^2 = (44-c^2)(21-2\sqrt{110})

c^2 = 924 - 88\sqrt{110} - c^2(21 - 2\sqrt{110})

c^2 + (21 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}

(22 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}

c^2 = \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}

c=\sqrt{ \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}}\approx 1.012\in (0, 2)

Hence, the function f(x) satisfies the Mean Value Theorem

7 0
4 years ago
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