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saul85 [17]
3 years ago
13

What is the true solution to the equation below? 2 In e^In2x -In e^In10x = In 30

Mathematics
2 answers:
stealth61 [152]3 years ago
8 0

The\ domain.\\D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\ \ \ |use\ \log_ab^n=n\log_ab\\\\2\ln2x\ln e-\ln10x\ln e=\ln30\ \ \ \ |use\ \log_aa=1\\\\2\ln2x-\ln10x=\ln30\ \ \ |use\ n\log_ab^n=\log_ab^n\\\\\ln(2x)^2-\ln10x=\ln30\\\\\ln4x^2-\ln10x=\ln30\ \ \ |use\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\ln\dfrac{4x^2}{10x}=\ln30\\\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2x}{5}=30\ \ \ \ |\cdot5\\\\2x=150\ \ \ \ |:2\\\\x=75\in D

Answer: x = 75

andrey2020 [161]3 years ago
6 0

Answer:

75

Step-by-step explanation:

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What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?
ollegr [7]

Answer:

25.6 units

Step-by-step explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

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

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

6 0
4 years ago
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Lelechka [254]
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Hope this helps and have a wonderful day!
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Answer:

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Step-by-step explanation:

output is always y

input is always x

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