Part 1 - Multiple Choice Direction: From the Option Letter A To D, Circle the Letter of the Options That Best…

Question

Part 1 - Multiple Choice Direction: From the Option Letter A To D, Circle the Letter of the Options That Best Answer the Question.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the problems:

1. If $f(x) = x^2 - 4x + 7$ , the minimum value of $f(x)$ is:

For a quadratic function $f(x) = ax^2 + bx + c$ with $a > 0$ , the minimum value occurs at the vertex.
The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$ .
For $f(x) = x^2 - 4x + 7$ , we have $a=1$ and $b=-4$ .
$x = -\frac{-4}{2(1)} = \frac{4}{2} = 2$
Substitute the x-coordinate of the vertex into the function to find the minimum value.
$f(2) = (2)^2 - 4(2) + 7$ $f(2) = 4 - 8 + 7$ $f(2) = 3$
The minimum value is 3.

The correct option is C.
$\boxed{\text{C. 3}}$
2. Solve: $\log_2(x-1) + \log_2(x+1) = 3$
Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ to combine the terms on the left side.
$\log_2((x-1)(x+1)) = 3$ $\log_2(x^2 - 1) = 3$
Convert the logarithmic equation to an exponential equation using the definition $\log_b M = N \iff b^N = M$ .
$2^3 = x^2 - 1$ $8 = x^2 - 1$
Solve for $x$ .
$x^2 = 8 + 1$ $x^2 = 9$ $x = \pm\sqrt{9}$ $x = \pm 3$
Check for valid solutions.
The arguments of logarithms must be positive.
For $x=3$ : $x-1 = 2 > 0$ and $x+1 = 4 > 0$ . So, $x=3$ is a valid solution.
For $x=-3$ : $x-1 = -4 < 0$ . So, $x=-3$ is not a valid solution.

The only valid solution is $x=3$ .

The correct option is B.
$\boxed{\text{B. 3}}$
3. The derivative of $x^x$ is:
Let $y = x^x$ .
Use logarithmic differentiation by taking the natural logarithm of both sides.
$\ln y = \ln(x^x)$ $\ln y = x \ln x$
Differentiate both sides with respect to $x$ .
Use the chain rule on the left side and the product rule on the right side.
$\frac{1}{y} \frac{dy}{dx} = (1)(\ln x) + (x)\left(\frac{1}{x}\right)$ $\frac{1}{y} \frac{dy}{dx} = \ln x + 1$
Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$ .
Substitute $y = x^x$ back into the equation.
$\frac{dy}{dx} = y(\ln x + 1)$ $\frac{dy}{dx} = x^x(\ln x + 1)$
The correct option is B.
$\boxed{\text{B. } x^x(\ln x + 1)}$
4. A geometric sequence has first term 3 and ratio 2. Sum of first 6 terms:
Identify the given values for the geometric sequence
first term $a = 3$ , common ratio $r = 2$ , and number of terms $n = 6$ .
Use the formula for the sum of the first $n$ terms of a geometric sequence
$S_n = \frac{a(r^n - 1)}{r-1}$ .
$S_6 = \frac{3(2^6 - 1)}{2-1}$
Calculate the sum.
$S_6 = \frac{3(64 - 1)}{1}$ $S_6 = 3(63)$ $S_6 = 189$
The correct option is A.
$\boxed{\text{A. 189}}$
5. If $\sin \theta = 3/5$ and $\theta$ is acute, $\cos \theta =$
Since $\theta$ is an acute angle, it is in the first quadrant, where both sine and cosine are positive.
Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ .
$\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$
Solve for $\cos^2 \theta$ .
$\frac{9}{25} + \cos^2 \theta = 1$ $\cos^2 \theta = 1 - \frac{9}{25}$ $\cos^2 \theta = \frac{25}{25} - \frac{9}{25}$ $\cos^2 \theta = \frac{16}{25}$
Take the square root to find $\cos \theta$ .
Since $\theta$ is acute, $\cos \theta$ must be positive.
$\cos \theta = \sqrt{\frac{16}{25}}$ $\cos \theta = \frac{4}{5}$
The correct option is A.
$\boxed{\text{A. } \frac{4}{5}}$
6. The solution of $|2x-5|=7$ is:
For an absolute value equation $|A|=B$ , there are two possibilities
$A=B$ or $A=-B$ .
So, we set up two equations:
1. $2x - 5 = 7$
2. $2x - 5 = -7$
Solve the first equation.
$2x - 5 = 7$ $2x = 7 + 5$ $2x = 12$ $x = \frac{12}{2}$ $x = 6$
Solve the second equation.
$2x - 5 = -7$ $2x = -7 + 5$ $2x = -2$ $x = \frac{-2}{2}$ $x = -1$
The solutions are $x=6$ and $x=-1$ .

The correct option is A.
$\boxed{\text{A. } 6, -1}$
7. The determinant of $[[1,2],[3,4]]$ is:
For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is given by the formula $ad - bc$ .
For the given matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , we have $a=1$ , $b=2$ , $c=3$ , and $d=4$ .
Calculate the determinant.
$\text{det} = (1)(4) - (2)(3)$ $\text{det} = 4 - 6$ $\text{det} = -2$
The correct option is A.
$\boxed{\text{A. -2}}$
8. $\lim_{x \to 0} \frac{\sin x}{x} =$
This is a fundamental limit in calculus.
Recall the standard limit.
$\lim_{x \to 0} \frac{\sin x}{x} = 1$
The correct option is B.
$\boxed{\text{B. 1}}$
9. If $x^2-5x+6=0$ , then $x$ equals:
This is a quadratic equation.
We can solve it by factoring. Look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.
$(x-2)(x-3) = 0$
Set each factor equal to zero and solve for $x$ .
$x - 2 = 0 \implies x = 2$ $x - 3 = 0 \implies x = 3$
The solutions are $x=2$ or $x=3$ .

The correct option is A.
$\boxed{\text{A. 2 or 3}}$

Part 1 - Multiple Choice Direction: From the Option Letter A To D, Circle the Letter of the Options That Best Answer the Question.

Answer

Step-by-step solution

For a quadratic function $f(x) = ax^2 + bx + c$ with $a > 0$ , the minimum value occurs at the vertex.

Substitute the x-coordinate of the vertex into the function to find the minimum value.

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ to combine the terms on the left side.

Convert the logarithmic equation to an exponential equation using the definition $\log_b M = N \iff b^N = M$ .

Solve for $x$ .

Check for valid solutions.

Let $y = x^x$ .

Differentiate both sides with respect to $x$ .

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$ .

Identify the given values for the geometric sequence

Use the formula for the sum of the first $n$ terms of a geometric sequence

Calculate the sum.

Since $\theta$ is an acute angle, it is in the first quadrant, where both sine and cosine are positive.

Solve for $\cos^2 \theta$ .

Take the square root to find $\cos \theta$ .

For an absolute value equation $|A|=B$ , there are two possibilities

Solve the first equation.

Solve the second equation.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is given by the formula $ad - bc$ .

Calculate the determinant.

This is a fundamental limit in calculus.

Recall the standard limit.

This is a quadratic equation.

Set each factor equal to zero and solve for $x$ .

Try one yourself

Still stuck on this one?

More Mathematics Questions

Part 1 - Multiple Choice Direction: From the Option Letter A To D, Circle the Letter of the Options That Best Answer the Question.

Answer

Step-by-step solution

For a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c with a>0a > 0a>0, the minimum value occurs at the vertex.

Substitute the x-coordinate of the vertex into the function to find the minimum value.

Use the logarithm property log⁡bM+log⁡bN=log⁡b(MN)\log_b M + \log_b N = \log_b (MN)logb​M+logb​N=logb​(MN) to combine the terms on the left side.

Convert the logarithmic equation to an exponential equation using the definition log⁡bM=N ⟺ bN=M\log_b M = N \iff b^N = Mlogb​M=N⟺bN=M.

Solve for xxx.

Check for valid solutions.

Let y=xxy = x^xy=xx.

Differentiate both sides with respect to xxx.

Solve for dydx\frac{dy}{dx}dxdy​ by multiplying both sides by yyy.

Identify the given values for the geometric sequence

Use the formula for the sum of the first nnn terms of a geometric sequence

Calculate the sum.

Since θ\thetaθ is an acute angle, it is in the first quadrant, where both sine and cosine are positive.

Solve for cos⁡2θ\cos^2 \thetacos2θ.

Take the square root to find cos⁡θ\cos \thetacosθ.

For an absolute value equation ∣A∣=B|A|=B∣A∣=B, there are two possibilities

Solve the first equation.

Solve the second equation.

For a 2×22 \times 22×2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[ac​bd​], the determinant is given by the formula ad−bcad - bcad−bc.

Calculate the determinant.

This is a fundamental limit in calculus.

Recall the standard limit.

This is a quadratic equation.

Set each factor equal to zero and solve for xxx.

Try one yourself

Still stuck on this one?

More Mathematics Questions

For a quadratic function $f(x) = ax^2 + bx + c$ with $a > 0$ , the minimum value occurs at the vertex.

Use the logarithm property $\log_b M + \log_b N = \log_b (MN)$ to combine the terms on the left side.

Convert the logarithmic equation to an exponential equation using the definition $\log_b M = N \iff b^N = M$ .

Solve for $x$ .

Let $y = x^x$ .

Differentiate both sides with respect to $x$ .

Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$ .

Use the formula for the sum of the first $n$ terms of a geometric sequence

Since $\theta$ is an acute angle, it is in the first quadrant, where both sine and cosine are positive.

Solve for $\cos^2 \theta$ .

Take the square root to find $\cos \theta$ .

For an absolute value equation $|A|=B$ , there are two possibilities

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the determinant is given by the formula $ad - bc$ .

Set each factor equal to zero and solve for $x$ .