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Marianna [84]
3 years ago
14

My Final Question! 50 points and Brainliest!

id="TexFormula1" title="ln(x-24)+ln(x+100)=3 ln 5" alt="ln(x-24)+ln(x+100)=3 ln 5" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Ne4ueva [31]3 years ago
4 0

Answer:

lnx-24ln+lnx+100ln=3ln5

2lnx+76ln=3ln5

jok3333 [9.3K]3 years ago
4 0
In(x-24)*(x+100)= In(5^3)

Multiply the parentheses

In(x^2+100x-24x-2400)= In(5^3)

Set the arguments equal

x^2-100x-24x-2400=5^3

Collect like terms, evaluate the power and move the constant to the left

x^2+76x-2525= 0

solve via quadratic equation

x= -76*square root of 76^2-4*2525 divided by 2*1

From that you get

-76+ or -126 divided by 2

The two solutions would be -101 or 25

but, the only answer would be 25 as -101 isn't in the defined range, and would make the statement negative


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Plz help
frozen [14]

Answer:

a. Required system of equations is:

c+p = 10

5c+4p = 46

b. 6 cakes were sold

Step-by-step explanation:

Let p be the number of pies and c be the number of cakes.

Then according to given statement " the store sold 10 baked goods"

c+p = 10\ \ \ Eqn\ 1

And

"A cake costs $5 and a pie costs $4"

5c+4p = 46\ \ \ \ Eqn\ 2

Using equation 1,

c=10-p

Putting this value of c in equation 2:

5(10-p)+4p = 46\\50-5p+4p = 46\\-p = 46-50\\-p = -4\\p = 4

Putting p = 4 in equation 1

c+4 = 10\\c = 10-4\\c =6

Hence,

a. Required system of equations is:

c+p = 10

5c+4p = 46

b. 6 cakes were sold

7 0
3 years ago
Please help im begging you
Sergio [31]

Answer:

the domain is ALL reals numbers except ZERO

- ∞ < x < 0    ∪   0 <  x < ∞

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Which functions can generate all the values in table 2​
Whitepunk [10]

Answer:

animania-1.12.2-1.6.2.jar

4 0
2 years ago
HELP HELP HELP PLS PLS IM CONFUSED
yuradex [85]

Answer:

y=19 z=19

Step-by-step explanation:

45 45 90 triangles are this: sides=hyp/sqrt(2)

mathXL lol

4 0
1 year ago
Which of the following is not one of the 8th roots of unity?
Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1.  Since z=1=1+0\cdot i, then

Rez=1,\\ \\Im z=0

and

|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.

This gives you \varphi=0.

Thus,

z=1\cdot(\cos 0+i\sin 0).

2. The 8th roots can be calculated using following formula:

\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.

Now

at k=0,  z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;

at k=1,  z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=2,  z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;

at k=3,  z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};

at k=4,  z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;

at k=5,  z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

at k=6,  z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;

at k=7,  z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};

The 8th roots are

\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.

Option C is icncorrect.

5 0
2 years ago
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