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alexdok [17]
3 years ago
9

Will give brainliest Determine the following values!!!!!

Mathematics
1 answer:
yan [13]3 years ago
5 0

Answer:

x = 26    m∠DAB =  80°      m∠ADC = 100°

Step-by-step explanation:

Given,

ABCD is a parallelogram in which AB ║ DC and AD║ BC.

m∠DAB = 4x-24  

m∠ADC = 2x+48

Solution,

Since ABCD is a parallelogram so sum of two consecutive angle is 180°.

m\angle DAB+m\angle ADC=180\°\\4x-24+2x+48=180\°\\6x+24=180\°\\6x=180-24\\6x=156\\x=\frac{156}{6}=26

Now substituting the value of x we get the value of ∠DAB and ∠ADC .

m\angle DAB=4x-24=4\times26-24=104-24=80\°

m\angle ADC=2x+48=2\times26+48=52+48=100\°

Thus the value of x is 26 and m∠DAB is 80° and m∠ADC is 100°

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The square of the sum of two numbers is 144 and the sum of of their square is 180, find the number... Please I really need the a
mr Goodwill [35]

Answer:

The numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3

Step-by-step explanation:

Let the two numbers be 'x' and 'y'.

Given:

The squares of the sum of two numbers = 144

The sum of their squares = 180

The sum of the numbers = x+y

The square of the sum of numbers = (x+y)^2

Square of first number = x^2

Square of second number = y^2

The sum of their squares = x^2+y^2

Now, as per question:

(x+y)^2=144---(1)\\\\x^2+y^2=180---(2)

Expanding equation (1) using the formula (a+b)^2=a^2+b^2+2ab

This gives,

(x+y)^2=144\\\\x^2+y^2+2xy=144\\\\\textrm{Plug in the value of }x^2+y^2\textrm{ from equation (2)}\\\\180+2xy=144\\\\2xy=144-180\\\\2xy=-36\\\\xy=-\frac{36}{2}\\\\xy=-18\\\\y=-\frac{18}{x}---(3)

Plug in the value of 'y' from equation (3) in equation (1). This gives,

x^2+(-\frac{18}{x})^2=180\\\\x^2+\frac{324}{x^2}=180\\\\x^4+324=180x^2\\\\x^4-180x^2+324=0\\\\\textrm{On solving, we get:}\\\\x^2=178.2\ or\ x^2=1.8\\\\\textrm{Square root both sides, we get:}\\\\x=\pm13.35\ or\ x=\pm1.35

Therefore,

y=-\frac{18}{x}=-\frac{18}{\pm13.35} = \pm1.35\ or\\\\y=-\frac{18}{x}=-\frac{18}{\pm1.35}=\pm13.3

Therefore, the numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3

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