Answer:
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>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2
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In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28
0
and ∠QRT=65
0
, then find the values of x and y.
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Solution
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Given, PQ⊥PS,PQ∥SR,∠SQR=28
∘
,∠QRT=65
∘
According to the question,
x+∠SQR=∠QRT (Alternate angles as QR is transversal.)
⇒x+28
∘
=65
∘
⇒x=37
∘
Also ∠QSR=x
⇒∠QSR=37
∘
Also ∠QRS+∠QRT=180
∘
(Linear pair)
⇒∠QRS+65
∘
=180
∘
⇒∠QRS=115
∘
Now, ∠P+∠Q+∠R+∠S=360
∘
(Sum of the angles in a quadrilateral.)
⇒90
∘
+65
∘
+115
∘
+∠S=360
∘
⇒270
∘
+y+∠QSR=360
∘
⇒270
∘
+y+37
∘
=360
∘
⇒307
∘
+y=360
∘
⇒y=53
∘
Step-by-step explanation:
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Let's find out how much she spent every month.
4000 (starting money) - 2800 (remaining money) = 1200 spent over 3 months
1200/3 = 400 per month was spent
So if she continues to spend 400 a month?
How many months are left? 12 (months of the year) - 3 (months she already spent) = 9
So 9 (remaining months) * 400 (amt per month) = 3600 she'll spend at the going rate over 9 months.
But she only has 2800 left.
2800 (remaining) - 3600 (estimated total of spending) = -800
So she will be 800$ in debt at the end of the year at the current rate.
Answer:
Step-by-step explanation:
Simplify the expression -4x^2(3x − 7).
A. -12x^3 + 28
<h2>B. -12x^3 + 28x^2<= your answer</h2>
C. -12x^3 − 28
D. -12x^3 − 28x^2
The question is:
Check whether the function:
y = [cos(2x)]/x
is a solution of
xy' + y = -2sin(2x)
with the initial condition y(π/4) = 0
Answer:
To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.
Let us do that.
y = [cos(2x)]/x
y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]
Now,
xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x
= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)
= -2sin(2x)
Which is the right hand side of the differential equation.
Hence, y is a solution to the differential equation.
