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2 years ago

Find the values of x and y

Mathematics

1 answer:

tino4ka555 [31]2 years ago

6 0

Answer:

You can find out x and y by solving these equations simultaneously.

First case:

according to the opposite angle theorem, (18y+5)° = (10x-61)°

Now let's simplify them;

18y-10x = -66

y= (10x-66) ÷ 18......................(i)

Second case:

Around that centre, it is always 360°.

And the angle opposite to the (x+10)v is also (x+10)°.

And sum of these 4 angles is 360°.

(18y+5)° + ( 10x-61)° + (x +10)° + (x +10)°= 360°

18y + 12x -36 = 360°

18y + 12x = 396...........................(ii)

Now replace the value of y from (i) into (ii):

18 ((10x-66) ÷ 18) +12x =396

(10x-66) +12x = 396

22x -66 = 396

22x = 462

x = 21

Now let's put this value of x into our equation (i):

y= (10(21)-66) ÷ 18

y= 8

Cheers :)

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From 10:00 am to noon, the rise in temperature is 4 degrees C, and from 3:00 pm to 5:00 pm, the fall in temperature is 6 degrees

lyudmila [28]

Answer:

(a) 2 degree celsius per hour

(b) 3 degree celsius per hour.

Step-by-step explanation:

From the question,

(a) Avearge rate of rise in temperature per hour(T') = Rise in temperature(T)/Time required for the rise(t).

T' = T/t.................... Euqation 1

Rise in temperature (T) = 4 degree celsius, t = 10.00 am to noon = 2 hours.

Substitute into equation 1

T' = 4/2

T' = 2 degree celsius per hour.

(b) Also,

Average rate of fall in temperature per hour (T') = Fall in temperature (T)/Time(t)

T' = T/t

Given: T = 6 degree celsius, t = 3:00 pm to 5:00 pm = 2 hours

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T' = 3 degree celsius per hour

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2 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

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3 years ago

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Answer:

The answer to your question is below

Step-by-step explanation:

Question 1

x = 5 Equation l

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y = 0

- Solution (5, 0)

Question 2

x + 16y = 20 Equation l

x = 4y Equation ll

Substitute equation ll in equation l

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y = 20/20

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x = 4(1)

x = 4

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2x + 8y = 20 Equation l

x = 2 Equation ll

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4 + 8y = 20

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