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Wittaler [7]
3 years ago
13

Which of the following equations represents a line that is perpendicular to y=3x+11 and intercepts the line at the point, (-2, 5

)?

Mathematics
1 answer:
yaroslaw [1]3 years ago
7 0
If the line is perpendicular the slope is the reciprocal so it would be
Y=-1/3x
Then if the line has to go through the point (-2,5) the y-intercept must be 5 making the answer
Y=-1/3x+5
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One of our brainliest, Konrad509, made this:
Reil [10]

\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x

A=10, B=11, C=12, etc.

\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2

121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that x\in\mathbb{N}. Moreover we can deduce that x\geq19 since there is J and J=19.

After you succesfully solve it, you should get the answer x=20.

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3 years ago
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kozerog [31]
50-60-79_%^*^%>?> ggg
4 0
2 years ago
X = 10<br><br> x = 2<br><br> x = two divided by five.<br><br> x = 11
KengaRu [80]

Answer:

x = 10

Step-by-step explanation:

You can try the answers to see which works. (The first one does.)

Or, you can solve for the variable:

Divide by 75

... (1/5)^(x/5) = 3/75 = 1/25

Recognize that 25 = 5^2, so ...

... (1/5)^(x/5) = (1/5)^2

Equating exponents, you have

... x/5 = 2

... x = 10 . . . . . multiply by 5

_____

You can also start by taking logarithms:

... log(75) +(x/5)log(1/5) = log(3)

... (x/5)log(1/5) = log(3) -log(75) = log(3/75) = log(1/25) . . . . simplify the log

... x/5 = log(1/25)/log(1/5) = 2 . . . . . simplify (or evaluate) the log expression

... x = 10 . . . . . multiply by 5

_____

"Equating exponents" is essentially the same as taking logarithms.

5 0
3 years ago
132 days to 125.4 days percent of change
Evgen [1.6K]

Answer:

132 days to 125.4 days percent of change

Step-by-step explanation:

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p= 125.4-132/132 x 100%

p= -6.6/132 x 100%

p= (-660/132%)

p= -5%

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Step-by-step explanation:

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