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3 years ago

[50 POINTS!!!] HELP! Solve This Logarithm. Find X. Please Type Steps!

Mathematics

2 answers:

Verdich [7]3 years ago

7 0

Answer:

Step-by-step explanation:

$\text{Domain:}\ x>0\\\\\log_83+\dfrac{1}{2}\log_825=\log_8x\qquad\text{use}\ n\log_ab=\log_ab^n\\\\\log_83+\log_825^\frac{1}{2}=\log_8x\qquad\text{use}\ a^\frac{1}{2}=\sqrt{a}\\\\\log_83+\log_8\sqrt{25}=\log_8x\\\\\log_83+\log_85=\log_8x\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log_8(3\cdot5)=\log_8x\\\\\log_815=\log_8x\iff15=x\to x=15\in D$

gogolik [260]3 years ago

3 0

Answer:

Your answer would be x = 15

I hope his helped and have a wonderful day!

~Real

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How to write a binomial expression in standard form that has a degree of 4

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(x+a)^4 = ^4C_0 x^4 + ^4C_1 x^{4-1} a + ^4C_2x^{4-2}a^2 + ^4C_3x^{4-3}a^3 + ^4C_4x^{4-4}a^4

= x^4 + 4x^3a + 6x^2a^2 + 4xa^3 + a^4

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4 years ago

Here all the other ones .

Crazy boy [7]

Answer:

The answer is b

Step-by-step explanation:

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3 years ago

Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.

jok3333 [9.3K]

Answer:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

$\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}$

And, the distance from P to (0, -2) will be:

$\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}$

So sum of the two distances must be 6. Therefore:

$d_1+d_2=6$

Now, by substitution:

$(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6$

Simplify. We can subtract the second term from the left:

$\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}$

Square both sides:

$(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)$

We can cancel the x² terms and continue squaring:

$y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4$

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

$-8y=36-12\sqrt{x^2+(y+2)^2}$

We can divide both sides by -4:

$2y=-9+3\sqrt{x^2+(y+2)^2}$

Adding 9 to both sides yields:

$2y+9=3\sqrt{x^2+(y+2)^2}$

And, we will square both sides one final time.

$4y^2+36y+81=9(x^2+(y^2+4y+4))$

Distribute:

$4y^2+36y+81=9x^2+9y^2+36y+36$

The 36y will cancel. So:

$4y^2+81=9x^2+9y^2+36$

Subtracting 4y² and 36 from both sides yields:

$9x^2+5y^2=45$

And dividing both sides by 45 produces:

$\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1$

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

5 0

3 years ago

Read 2 more answers

Answer correctly for brainliest

mr_godi [17]

Answer:

x=3

Step-by-step explanation:

When you substitute x = 3 to both equations, you get y = -4 for both so that is a solution

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3 years ago

Which of the following functions is graphed below

elena-s [515]

Answer:

Step-by-step explanation:

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3 years ago