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Rzqust [24]
3 years ago
9

What is 4^2+8x-11x+6-5x^2+2​

Mathematics
1 answer:
Lelechka [254]3 years ago
8 0
<h3>I'll teach you how to solve 4^2+8x-11x+6-5x^2+2​</h3>

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

4^2+8x-11x+6-5x^2+2​

Add similar elements:

8x - 11x= -3x

4^2-3x+6-5x^2+2

Group like terms:

-5x^2-3x+4^2+6+2

Add the numbers:

6+2= 8

-5x^2-3x+4^2+8

4^2= 16

-5x^2-3x+16+8

Add the numbers:

16+8= 24

-5x^2-3x+24

Your Answer Is -5x^2-3x+24

plz mark me as brainliest if this helped :)

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When Janice is 20 years old; her brother is half her age. When Janice is 30 years old; how old is her brother?
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The foot of a ladder is placed 7meters from a wall. If the top of the ladder rests ,How long is the ladder
Virty [35]

Answer: 11.4 m

Step-by-step explanation:

The scenario depicted is that of a right angled triangle with the 7 meters being the base and the 9 meters being the height.

The length of the ladder will therefore be the Hypothenuse.

Using the Pythagorean Rule:

a² + b² = c²

with c being the hypothenuse

7² + 9² = c²

49 + 81 = c²

c²= 130

c = √130

c = 11.4 m

6 0
2 years ago
For which function is the domain -5 ≤ x ≤ -1 not appropriate? y = √3x y = 3x y = 3x y = x3
Maurinko [17]
There is no real values for the square root of a negative value, therefore that negative domain does not exist for the function y=√(3x).
4 0
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}

5 0
1 year ago
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