Mathematics Test 1 Sat Examination

Correct answer: B) The value of the bank account is estimated to be approximately 243 dollars in 1962.
f(x)=206(1.034)^x
where x is the number of years after 1957. 
So, f(5) represents the value of the account 5 years after 1957, which is the year 1962.
And f(5)≈243, meaning the value is approximately 243 dollars in 1962.

Original width = $w$  
Original length = $\frac{7}{2}w$  

New width = $w+7$

To maintain the same ratio:

$\text{New length}=\frac{7}{2}(w+7)=\frac{7}{2}w+\frac{7}{2}\cdot 7=\frac{7}{2}w+24.5$

So, the length must increase by 24.5 units to maintain the ratio.
 

The correct answer is A) f (x ) = (x + 44)²
​​​​​

Because:

Length of side P = L_P = x
Length of side P = L_Q
Perimeter of  P = P_P = 4L_P = 4x
Perimeter of  Q = P_Q =  4L_Q = 4x + 176
4L_Q = 4L_P + 176
L_Q = L_P + 44
​​​​​​But L_P + x
L_Q = x + 44
​​​​Area of  P = A_P
Area of  Q = A_Q
f(x) = Area Q, A_Q = L_Q²

Thus, 

f(x) = (x + 44)²

The correct answer is C. 

Given\[\dfrac{{14x}}{{7y}} = 2\sqrt {w + 19} \]\[\dfrac{x}{y} = \sqrt {w + 19} \]\[{\left( {\dfrac{x}{y}} \right)^2} = w + 19\]\[w = {\left( {\dfrac{x}{y}} \right)^2} - 19\]

We are given a triangle with side lengths:
$$a=2\sqrt{2},b=6\sqrt{2},c=\sqrt{80}$$
Simplify the square root
$$\sqrt{80}=\sqrt{16\times 5}=\sqrt{16}\cdot \sqrt{5}=4\sqrt{5}$$

So the sides become:
$$a=2\sqrt{2},b=6\sqrt{2},c=4\sqrt{5}$$

Check for a right triangle

We test whether:

$$a^2+b^2=c^2$$

Compute each side:

$$a^2=(2\sqrt{2}{)}^2=4\cdot 2=8$$
$$b^2=(6\sqrt{2}{)}^2=36\cdot 2=72$$
$$c^2=(4\sqrt{5}{)}^2=16\cdot 5=80$$

Since:

$$a^2+b^2=8+72=80=c^2$$

The triangle is a \textbf{right triangle}, with legs a and b, and hypotenuse c.
Use the right triangle area formula

$$\text{Area}=\frac{1}{2}ab$$

\[ = \dfrac{1}{2} \cdot 2\sqrt{2} \cdot 6\sqrt{2}
= \dfrac{1}{2} \cdot 12 \cdot (\sqrt{2})^2
= \dfrac{1}{2} \cdot 12 \cdot 2
= \dfrac{1}{2} \cdot 24
= 12\]

Given the expression:

\( 4x^2 + bx - 45 \)

In factored form:

\( (hx + k)(x + j) \)

Expand the factored form:

\[ (hx + k)(x + j) = hx(x + j) + k(x + j) = hx^2 + hjx + kx + kj \]

Collect like terms:

\[ hx^2 + (hj + k)x + kj \]

Match with the original expression:

\[ 4x^2 + bx - 45 \]

Matching coefficients:

• \( h = 4 \)
• \( kj = -45 \)
• \( hj + k = b \)

Identify which expression must be an integer:

Since \( h = 4 \) and \( kj = -45 \), and \( k \) and \( j \) are integers, their product \( kj = -45 \) must be an integer.

Conclusion:

\( kj \) must be an integer.

The corect answer is C) 6

Given
y = 2x² − 21x + 64    _____Eqn 1
y = 3x + a   ____________Eqn 2

At point of intersect:
2x² − 21x + 64 = 3x + a 
2x² − 21x + 64 −  (3x + a ) = 0
2x²− 24x + 64 - a = 0
This is a quadratic equation; And the  discriminat defined by
D = b² - 4ac = 0
a = 2 (i.e First term);
b = -24 (i.e Second term)
c = 64 - a (i.e Third term)
Subtituting these values into the discriminant equation; we have:
(− 24)²− 4(2)(64− a) = 0
576 − 8(64 − a) = 0
a = -8

Now subitituting a = -8 into the eqn 2
y = 3x − 8

Equating eqn 1 and 2 again
2x² − 21x + 64 = 3x − 8
2x² − 24x + 72 = 0
x² − 12x + 36 = 0
(x - 6)² = 0
x = 6

​

The correct option is C) 58 + 58√2
In an isosceles right triangle, the two legs are equal, and the angle between them is 90^o.
If the legs are x, then the hypotenuse is: x√2
x√2 = 58\[x=\dfrac{58}{\sqrt{2}}\]\[x=29\sqrt{2}\]

Perimeter= x + x + 58
= 2x + 58\[=2(29\sqrt{2})+58\]\[=58\sqrt{2}+58\]

The parabola has vertex (9, −14) and intersects the x-axis at two points, so it opens upward or downward and has real roots.
​​Use vertex form of a parabola:

y=a(x−h)²+k
where (h,k) is the vertex:
So, 
y = a(x - 9)² - 14
y = a(x² - 18x + 81) - 14
y = ax² - 18ax + 81a -14
comparing with y = ax² + bx + c
b = -18a, 
c = 81a - 14

Compute a + b + c 
a + b + c  = a + (-18a) + (81a -14) = 64a - 14
a + b + c = 64a - 14
Since the vertex of the parabola, (9, -14), is below the x-axis, and it’s given that the parabola intersects the x-axis at two points, the parabola must open upward.

Therefore, the constant a must have a positive value. Setting the expression 64a-14 equal to the value in choice D yields 64a-14=-12.
a = 2/64.
a = 1/32
Recall 
b = -18a, b = -18(1/32) = -9/16
c = 81a - 14 = 81(1/32) - 14 = -367/32

\[a+b+c\text{ }=~\dfrac{2}{64}-\dfrac{9}{16}-\dfrac{367}{32}=-12\]

Correct answer is B) 1994

From the graph, the year 1994 has the greatest number of chipmunks in the state park