Brenda is selling two different nut mixes as snacks. Mix A has 3 oz. of cashews and 7 oz. of almonds. Mix B has 6 oz. of cashews and 5 oz. of almonds. Which inequalities represent the number of mix A mixes, A, and mix B mixes, B, Brenda can make if she has 72 oz. of cashews and 84 oz. of almonds?
A. 3A+7B≤72
6A+5B≤84
B. 3A+6B≤72
7A+5B≤84
C. 3A+7B≤84
6A+5B≤72
D. 6A+3B≤72
5A+7B≤84
The amount of cashews in mix A and mix B should add up to 72 oz., and the amount of almonds in mix A and mix B should add up to 84 oz. This leads to the following system of inequalities:
3A + 6B ≤ 72 (3 oz. of cashews in mix A and 6 oz. of cashews in mix B)
7A + 5B ≤ 84 (7 oz. of almonds in mix A and 5 oz. of almonds in mix B)
Therefore, the answer is choice B: 3A+6B≤72 and 7A+5B≤84.
Simplify the expression 6x^4x^8/x^2+x^2
First, we can simplify the numerator by multiplying 6 and adding the exponents of x. This gives us:
6x^(4+8) = 6x^12
Now, in the denominator, we can combine like terms by factoring out x^2:
x^2 + x^2 = 2x^2
So, the expression simplifies to:
6x^12 / 2x^2
We can simplify this further by canceling out 2 and dividing the exponents of x:
3x^(12-2) = 3x^10
Therefore, the simplified expression is 3x^10.
5. The quantity of dollars in a bank account is given by the function f(x)=15(1.02)^,x where x is the number of years from the date at which money is initially deposited. Given this function, which of the following is true?
A. $1.02 were initially deposited into the account, which grows at an annual rate of 0.15%.
B. $15 were initially deposited into the account, which grows at an annual rate of 2%.
C. $15 were initially deposited into the account, which grows at an annual rate of 102%.
D. $1.02 were initially deposited into the account, which grows at an annual rate of 15%
6. 50 ng of a radioactive isotope with a half-life of 23 seconds is expected to decay exponentially. How much of the substance will remain after 10 seconds? Round your answer to the nearest tenth of a nanogram.
A. 0.4 ng
B. 20.0 ng
C. 0.7 ng
D. 37.0 ng
7. A table of values for the exponential function f(x) is shown. What is the exponential function that it represents?
X f(x)
0 6
1 18
2 54
3 162
A. f(x)=6(3) ^x
B. f(x)=6(2)^x
C. f(x)=4(3)^x
D. f(x)=3(6)^x
9. The graph of f(x) is given above. Which of the following functions increase or decrease as f(x) does over (−∞, 3) and (3, ∞)? Select the two correct answers.
A. g(x)=|x−3|
B. g(x)=−(x−3)^2
C. g(x)=−|x−3|
D. g(x)=(x−3)^2
E. g(x)=x−3
10. A function is given such that f(4)−f(6)=9. If the function is linear, which of the following could the function possibly be?
A. f(x)=4.5x+b for some constant b.
B. f(x)=−4.5x+b for some constant b
C. f(x)=−9x+b for some constant b.
D. f(x)=9x+b for some constant b.
11. The values of a linear function are given in the table below:
What is the point with a positive x-coordinate at which the exponential function g(x)=3x exceeds f(x)?
A. g(x) and f(x) have the same y-value at (9, 2) since f(9)=2, g(9)=2, and g(x)>f(x)>9 for all x>2.
B. g(x) and f(x) have the same y-value at (2, 9) since f(2)=9, g(2)=9, and g(x)>f(x)>9 for all x>2.
C. g(x) and f(x) have the same y-value at (9, 2) since f(9)=2, g(9)=2, and g(x)>f(x)>2 for all x>9.
D. g(x) and f(x) have the same y-value at (2, 9) since f(2)=9, g(2)=9, and g(x)>f(x)>2 for all x>9.
5. B. $15 were initially deposited into the account, which grows at an annual rate of 2%.
6. A. 0.4 ng (Note: The formula for exponential decay is A = A0(1/2)^(t/T), where A is the final amount, A0 is the initial amount, t is the time elapsed, and T is the half-life. Plugging in the values given, we get A = 50(1/2)^(10/23) ≈ 0.4 ng. )
7. B. f(x)=6(2)^x
9. B. g(x)=−(x−3)^2 and D. g(x)=(x−3)^2 (Note: Both of these functions are quadratic, which means they increase on one side of their vertex and decrease on the other side. The vertex of g(x) is at x = 3, so it matches the behavior of f(x) over both intervals. Therefore, g(x) can either be the upward facing parabola or the downward facing parabola centered at x = 3.)
10. D. f(x)=9x+b for some constant b. (Note: Since f(4)−f(6)=9, we know that the slope of the function is -9/2. One possible formula for a linear function with this slope is f(x)=-9x/2+b, which simplifies to f(x)=9x/2+b if we let b be any constant.)
11. B. g(x) and f(x) have the same y-value at (2, 9) since f(2)=9, g(2)=9, and g(x)>f(x)>9 for all x>2. (Note: We can solve for x in the inequality g(x) > f(x) and get x > log(3/2)/log(3). This tells us that the functions first intersect when g(x) and f(x) have the same y-value somewhere to the right of x = log(3/2)/log(3), which is between 2 and 3. We can plug in x = 2 to both functions and see that they have the same y-value of 9, so this is the correct answer.)
△XYZ is given with X(2, 0), Y(0, −2), and Z(−1, 1). Which of the following can be used to prove that △XYZ is isosceles?(1 point)
To solve this question, we need to set up a system of inequalities based on the given constraints. Let's first determine the number of Mix A mixes (A) and Mix B mixes (B) that Brenda can make.
For the cashews, if Mix A has 3 oz. and Mix B has 6 oz., the total cashews used will be 3A + 6B.
Similarly, for the almonds, if Mix A has 7 oz. and Mix B has 5 oz., the total almonds used will be 7A + 5B.
The constraints given in the problem are that Brenda has 72 oz. of cashews and 84 oz. of almonds. Thus, we can set up the following inequalities:
3A + 6B ≤ 72 (equation 1) - This inequality represents the constraint for the cashews.
7A + 5B ≤ 84 (equation 2) - This inequality represents the constraint for the almonds.
Looking at the answer choices, we can see that the correct option is B: 3A + 6B ≤ 72 and 7A + 5B ≤ 84.
Therefore, Brenda can make a number of Mix A mixes (A) and a number of Mix B mixes (B) that satisfy these inequalities.