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3 years ago

Find the two inequalities that define the region on the grid that is not shaded

Mathematics

1 answer:

hoa [83]3 years ago

3 0

Answer:

y 》 2

y > -x + 3

Step-by-step explanation:

Equation of the bold line:

m = (-3/3) = -1

y = -x + 3

Shaded region is below y=2 (dotted)

So, it's defined by y < 2

Also, it's below the bold line

y 《 - x + 3

Everything outside these is not shaded:

y 》2

and y > -x + 3

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Given the functions f(x) = 7x + 13 and g(x) = x + 2, which of the following functions represents f[g(x)] correctly? (2 points)

Svetach [21]

Answer:

The value of f[ g(x) ] = 7x + 27

Step-by-step explanation:

It is given that, f( x ) = 7x + 13 and g( x ) = x + 2

<u>To find the value of f(g(x))</u>

g(x) = x + 2 and 7x + 13 (given)

Let g(x) = x + 2

f [ g(x) ] = 7(x + 2) + 13 [ substitute the value of g(x) in f(x) ]

= 7x + 14 + 13

= 7x + 27

Therefore the value of f[ g(x) ] = 7x + 27

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2 years ago

Round the following numbers to the nearest 1000

Fed [463]

Answer:

A. 5000 B: 12000 C 57000

Step-by-step explanation:

All you need to do is look at the hundreds place number and if it is below 4 or is 4 keep the number the same, if it is above 4, increase it by one.

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3 years ago

Read 2 more answers

Find the value of x for which p is parallel to q, if m∠1 = (9x) and m∠3 = 117.

kvasek [131]

If p||q then, m∠1=m∠3

9x = 117
x = 117/9
x = 13

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3 years ago

Read 2 more answers

A golfers score after playing on friday was +2

jek_recluse [69]

The end of his round on Saturday was a -5, and the end of his round on Sunday was a 0.

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2 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago