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Ghella [55]
2 years ago
8

Can someone help? I'm awful at this :/

Mathematics
2 answers:
tester [92]2 years ago
7 0
What do you need help with????
Nat2105 [25]2 years ago
5 0

Answer:

help with what

Step-by-step explanation:

life........ dont we all

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7x + 4 = 7x + 4
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Infinitely many because if you subtract 7x from both sides, you get 4 = 4, which is true at all times no matter what x is.

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I need help with #28-31 please
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(28) 2x+10+x-4=21
(29) 3x-16+4x-8=60
(30) 2x-8+3x-10=17
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In a mapping diagram each _____ corresponds to an ordered pair. A.) input value B.)relation C.)arrow D.)output value
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The answer is c. relation
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3 years ago
If <img src="https://tex.z-dn.net/?f=x%20%3D%209%20-%204%5Csqrt%7B5%7D" id="TexFormula1" title="x = 9 - 4\sqrt{5}" alt="x = 9 -
Komok [63]

Observe that

\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = \left(\sqrt x\right)^2 - 2\sqrt x\dfrac1{\sqrt x} + \left(\dfrac1{\sqrt x}\right)^2 = x - 2 + \dfrac1x

Now,

x = 9 - 4\sqrt5 \implies \dfrac1x = \dfrac1{9-4\sqrt5} = \dfrac{9 + 4\sqrt5}{9^2 - \left(4\sqrt5\right)^2} = 9 + 4\sqrt5

so that

\left(\sqrt x - \dfrac1{\sqrt x}\right)^2 = (9 - 4\sqrt5) - 2 + (9 + 4\sqrt5) = 16

\implies \sqrt x - \dfrac1{\sqrt x} = \pm\sqrt{16} = \pm 4

To decide which is the correct value, we need to examine the sign of \sqrt x - \frac1{\sqrt x}. It evaluates to 0 if

\sqrt x = \dfrac1{\sqrt x} \implies x = 1

We have

9 - 4\sqrt5 = \sqrt{81} - \sqrt{16\cdot5} = \sqrt{81} - \sqrt{80} > 0

Also,

\sqrt{81} - \sqrt{64} = 9 - 8 = 1

and \sqrt x increases as x increases, which means

0 < 9 - 4\sqrt5 < 1

Therefore for all 0 < x < 1,

\sqrt x - \dfrac1{\sqrt x} < 0

For example, when x=\frac14, we get

\sqrt{\dfrac14} - \dfrac1{\sqrt{\frac14}} = \dfrac1{\sqrt4} - \sqrt4 = \dfrac12 - 2 = -\dfrac32 < 0

Then the target expression has a negative sign at the given value of x :

x = 9-4\sqrt5 \implies \sqrt x - \dfrac1{\sqrt x} = \boxed{-4}

Alternatively, we can try simplifying \sqrt x by denesting the radical. Let a,b,c be non-zero integers (c>0) such that

\sqrt{9 - 4\sqrt5} = a + b\sqrt c

Note that the left side must be positive.

Taking squares on both sides gives

9 - 4\sqrt5 = a^2 + 2ab\sqrt c + b^2c

Let c=5 and ab=-2. Then

a^2+5b^2=9 \implies a^2 + 5\left(-\dfrac2a\right)^2 = 9 \\\\ \implies a^2 + \dfrac{20}{a^2} = 9 \\\\ \implies a^4 + 20 = 9a^2 \\\\ \implies a^4 - 9a^2 + 20 = 0 \\\\ \implies (a^2 - 4) (a^2 - 5) = 0 \\\\ \implies a^2 = 4 \text{ or } a^2 = 5

a^2 = 4 \implies 5b^2 = 5 \implies b^2 = 1

a^2 = 5 \implies 5b^2 = 4 \implies b^2 = \dfrac45

Only the first case leads to integer coefficients. Since ab=-2, one of a or b must be negative. We have

a^2 = 4 \implies a = 2 \text{ or } a = -2

Now if a=2, then b=-1, and

\sqrt{9 - 4\sqrt5} = 2 - \sqrt5

However, \sqrt5 > \sqrt4 = 2, so 2-\sqrt5 is negative, so we don't want this.

Instead, if a=-2, then b=1, and thus

\sqrt{9 - 4\sqrt5} = -2 + \sqrt5

Then our target expression evaluates to

\sqrt x - \dfrac1{\sqrt x} = -2 + \sqrt5 - \dfrac1{-2 + \sqrt5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - \dfrac{-2 - \sqrt5}{(-2)^2 - \left(\sqrt5\right)^2} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 + \dfrac{2 + \sqrt5}{4 - 5} \\\\ ~~~~~~~~~~~~ = -2 + \sqrt5 - (2 + \sqrt5) = \boxed{-4}

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