The solution to given system of equations is x = 2 and y = -5
<em><u>Solution:</u></em>
<em><u>Given the system of equations are:</u></em>
4x + y = 3 ---------- eqn 1
-2x + 3y = -19 ---------- eqn 2
We have to find the solution to above system of equations
<em><u>We can solve the system by substitution method</u></em>
From eqn 1,
4x + y = 3
Isolate y to one side
y = 3 - 4x ----------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
-2x + 3(3 - 4x) = -19
-2x + 9 - 12x = -19
Combine the like terms
-14x = -19 - 9
-14x = -28
Divide both sides of equation by -14
<h3>x = 2</h3>
<em><u>Substitute x = 2 in eqn 3</u></em>
y = 3 - 4(2)
y = 3 - 8
<h3>y = -5</h3>
Thus the solution is x = 2 and y = -5
<h3>
Answer: ds/dt = 11</h3>
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Work Shown:
Before we can use derivatives, we need to find the value of s when (x,y) = (15,20)
s^2 = x^2+y^2
s^2 = 15^2+20^2
s^2 = 225+400
s^2 = 625
s = sqrt(625)
s = 25
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Now we can apply the derivative to both sides to get the following. Don't forget to use the chain rule.
s^2 = x^2 + y^2
d/dt[s^2] = d/dt[x^2 + y^2]
d/dt[s^2] = d/dt[x^2] + d/dt[y^2]
2s*ds/dt = 2x*dx/dt + 2y*dy/dt
2(25)*ds/dt = 2(15)*5 + 2(20)*(10)
50*ds/dt = 150 + 400
50*ds/dt = 550
ds/dt = 550/50
ds/dt = 11
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Side note: The information t = 40 is never used. It's just extra info.
Given:
It is given that,
PQ ⊥ PS and
∠QPR = 7x-9
∠RPS = 4x+22
To find the value of ∠QPR.
Formula
As per the given problem PR lies between PQ and PS,
So,
∠QPR+∠RPS = 90°
Now,
Putting the values of ∠RPS and ∠QPR we get,
or, 
or, 
or, 
or, 
Substituting the value of in ∠QPR we get,
∠QPR = 
or, ∠QPR = 
Hence,
The value of ∠QPR is 40°.
Answer:
please check the attached file for the answer
Step-by-step explanation:
the explanation is in the attached file
