EmSAT Mathematics Mock Test-4 - Grade 8 MCQ

x² + 7x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x² + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x²)

Product of zeroes = 10 = Constant term / Coefficient of x²

The correct solution of this question is given below:

Here, P(a) = x²a² - 2xa + 3x - 2

1 is a zero of P(a), so P(1) = 0

Therefore, x²1² - 2x.1 + 3x - 2 = 0

x² + x - 2 = 0

(x + 2)(x - 1) = 0

x = -2, 1

Temperature is defined as the degree of hotness.
Temperature is measured using different scales in different regions.
we know that
 C = temperature in celcius.
F = temperature in Fahrenheit.
F= 9(C)/5 + 32
1) C = 30
F= 9(30)/5 + 32
F= 54 + 32 = 86F
2) C = 35
F= 9(35)/5 + 32
F= 63 + 32 = 95F
The range of temperatures in degree Fahrenheit is 86 F to 95 F.

The equation (a) 1/x + 1/y = 4 is not a linear equation because the variables x and y are in the denominator.

[√x]² + [√y]² = √2
Step 1: Simplify the terms
• [√x]² = x
• [√y]² = y
Thus, the equation becomes:
x + y = √2
Step 2: Check for linearity
• The variables  x and y both have degree 1.
• The constant √2 does not affect the linearity of the equation.
Conclusion:
This is a linear equation because it is of the form ax + by + c = 0, where a = 1, b = 1, and c = - √2.
Answer: The equation is linear.

The correct answer is a) A

• B' gives us all the elements in U other than 6 and 7 i.e., B' = {1, 2, 3, 4, 5, 8, 9, 10}
• The intersection of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is set A itself.

The Magnitude of y = 3sinx is 3.
graph{y=-3*sinx [-10, 10, -5, 5]}
Amplitude is the height of a periodic function, aka the distance from the center of the wave to it's highest point (or lowest point). You can also take the distance from the highest point to the lowest point of the graph and divide it by two.

count from rigth to left till you get last non-zero number 
like here 2 is the last non-zero number and for it we have to count 7 times
then write 2.16x 10⁷

y = 2783/253

y = 11

A and B are symmetric matrices, therefore, we have:
A′=A and B′=B..........(i)
 
Consider
(AB−BA)′=(AB)′ − (BA)′,[∵(A−B)′=A′B′]
 
=B′A′− A′B',[∵(AB)′= B′A]
 
=BA−AB [by (i) ]
 
=−(AB−BA)
 
∴(AB−BA) ′=−(AB−BA)
 
Thus, (AB−BA) is a skew-symmetric matrix.

We’ll first need to divide the function out and simplify before we take the derivative. Here is the rewritten function.

The derivative is, 

 has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.

and ∠D' = ∠D = 85°

∴ ∠C' + ∠D' = 70° + 85° = 155°

Area of triangle = 1/2 *base*height
Area of ABD=1/2 *BD*2
Area of  BDC=1/2*BD*3

Applying componendo and dividendo 
[sin(A+C/2) + sin (C/2)]/[sin(A+C/2) - sin (C/2)] = (k+1)/(k-1)
⇒ [2sin(A+C)cos A/2]/[2cos(A+C)sin A/2] ​= (k+1)/(k−1)
⇒ [tan(A+C)/2]/(tan A/2) ​= (k+1)/(k-1)
​⇒ [tan(π−B)/2]/(tan A/2) = (k+1)/(k-1)
​⇒ 1/(tan A/2 tan B/2) = (k+1)/(k-1)    
⇒ tan A/2 tan B/2 = (k-1)/(k+1)

Given the equation is,
2x²+3y²−8x−18y+35=K
Or, 2{x²−4x+4} + 3{y²−6y+9}=K
Or, 2(x−2)² + 3(y−3)² =K.
From the above equation it is clear that if K>0 then the given equation will represent an ellipse and for K<0, no geometrical interpretation.
Also if K=0 then the given equation will be reduced to a point and the point will be (2,3).

P(x,y) = [(e^t + e^-t)/2 , (e^t - e^-t)/2]
(e^t + e^-t)/2 = x --------------------------(1)
(e^t - e^-t)/2 = y --------------------------(2)
Adding (1) & (2)
2e^t = 2x + 2y
e^t = x + y
Eq (1) e^t + e^-t = 2x
e^t + 1/e^t = 2x
(e^t)² + 1 = 2x*e^t
(x+y)² + 1 = 2x(x+y)
x² + y² + 2xy + 1 = 2x² + 2xy
x² + y² + 1 = 2x²
(x²)/(1)² - (y²)/(1)² = 1  {which represents hyperbola equation} 

The curve y=(x)^1/2 is the part of curve y²=x
Equation of normal at P(t^2/4, t/2) is
y+tx=t/2+t3/4..... (1)
The equation will pass through (3,6)
6+3t=t/2+t3/4
t³−10t−24=0
Solving, we get t=4
So equation of line which is orthogonal and passes through (3,6) is
y+4x=18
4x+y-18 = 0

Equation of tangent and normal at P(at²,2at) on y²=4ax are
ty=x+at².....(1)
y+tx=2at+at³...(2)
So, T(−at²,0) & G(2a+at²,0) 
Equation of circle passing through P,T and G is
(x+at²)(x−(2a+at²))+(y−0)(y−0)=0
x²+y²−2ax−at²(2a+at²)=0
Equation of tangent on the above circle at P(at²,2at) is at²x+2aty−a(x+at²)−at²(2a+at²)=0
Slope of line which is tangent to circle at P

ρ = (b(xy) * b(yx))
But sign of ρρ is same as sign of b(xy), b(yx)
Therefore, ρ = 2/7

Since 1 ticket is choosen out of 2000 tickets
n(S) = ²⁰⁰⁰C₁
= (2000!/1!1999!)
= 2000
Now out of 2000 tickets only 50 have a prize
Hence no of tickets not having prize
= 2000 - 50
= 1950
Let A be the event that if we buy 1 ticket it doesnt have a prize
Hence, 1 ticket will be out of 1950 tickets
n(A) = ¹⁹⁵⁰C₁
= 1950
Probability not getting a prize if we get one ticket P(A) = n(A)/n(S) 
= 1950/2000