Trigonometric Ratios Quiz Answer Key
1. 0.8
2. 17.3
3. 36.9°
4. 23.6°
5. 18
6. 1
7. plus-or-minus 2√5
8. 2nd choice/(Image: A curve is graphed on a coordinate grid is shown. The curve starts at the point left-parenthesis negative 3 comma 0 right-parenthesis, moves up and to the right while curving slightly downward, and has an arrow on the end.)
9. 212 meters
10. 40°
100% correct!
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THANK YOU SO MUCH, CONNEXUS STUDENT!!! THEIR RIGHT!!
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THANKYOU SO MUCH I WAS ABOUT TO GET A 0 BUT YOU SAVED ME THANKYOU THANKYOU
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Bot I wasn't talking to you I was talking to Conexus student.
I'm sorry about the confusion. You're welcome nonetheless. If you need any help in the future, feel free to reach out!
The graph of `f\left(x\right)=x^{2}`was translated `2.5` units to the left to create the graph of function `g`. Which function represents `g`?
The function `g` can be represented as `g(x) = (x + 2.5)^2`. This is because moving `2.5` units to the left means subtracting `2.5` from `x`, so `g(x)` would be equal to `f(x - 2.5)`. We can substitute `f(x) = x^2` into this expression to get `g(x) = (x - 2.5)^2`, which can be simplified to `g(x) = (x + 2.5)^2` by expanding the square. Therefore, the function `g` is `g(x) = (x + 2.5)^2`.
Which answer choice describes how the graph of `f(x)=x^{2}` was transformed to create the graph of `h(x)=x^{2}-1`?
The graph of `h(x) = x^2 - 1` was created by shifting the graph of `f(x) = x^2` one unit down. This is because the `-1` in `h(x) = x^2 - 1` shifts the entire graph of `f(x) = x^2` down by one unit. Specifically, every point on the graph of `f(x)` gets shifted one unit downwards to create a corresponding point on the graph of `h(x)`. Therefore, the correct answer is "The graph was shifted down 1 unit."
The graph of `f\left(x\right)=x^{2}`is shown on the grid. Which statement about the relationship between the graph of `f`and the graph of `g(x)=5x^{2}` is true?
The graph of `g(x) = 5x^2` is an elongated version of the graph of `f(x) = x^2`. This is because `g(x) = 5x^2` stretches the graph of `f(x) = x^2` vertically by a factor of `5`, which makes the graph taller and narrower. More specifically, the `5` in front of `x^2` in `g(x)` stretches the graph vertically by a factor of `5`. Therefore, the correct answer is "The graph of `g(x) = 5x^2` is an elongation of the graph of `f(x) = x^2`."
Which value of `x` is a solution to this equation?
`2x^{2}+6x-20=0`
We can solve the equation `2x^2 + 6x - 20 = 0` by factoring it or by using the quadratic formula, but either way, we end up with two solutions:
`2x^2 + 6x - 20 = 0`
`2(x^2 + 3x - 10) = 0`
`2(x + 5)(x - 2) = 0`
`x + 5 = 0` or `x - 2 = 0`
`x = -5/2` or `x = 2`
Therefore, the solutions to the equation `2x^2 + 6x - 20 = 0` are `x = -5/2` and `x = 2`.
Solve `x^{2}-4x-5=0`
To solve `x^2 - 4x - 5 = 0`, we can use the quadratic formula:
`x = (-b ± sqrt(b^2 - 4ac)) / 2a`
where `a`, `b`, and `c` are the coefficients of the quadratic equation (`ax^2 + bx + c = 0`).
In this case, `a = 1`, `b = -4`, and `c = -5`, so we substitute these values into the formula:
`x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / 2(1)`
Simplifying the expression under the square root:
`x = (4 ± sqrt(16 + 20)) / 2`
`x = (4 ± sqrt(36)) / 2`
`x = (4 ± 6) / 2`
So the two solutions are:
`x = 5` or `x = -1`
Therefore, `x^2 - 4x - 5 = 0` when `x = 5` or `x = -1`.
Solve `2x^{2}-5x-3=0`
To solve `2x^2 - 5x - 3 = 0`, we can use the quadratic formula:
`x = (-b ± sqrt(b^2 - 4ac)) / 2a`
where `a`, `b`, and `c` are the coefficients of the quadratic equation (`ax^2 + bx + c = 0`).
In this case, `a = 2`, `b = -5`, and `c = -3`, so we substitute these values into the formula:
`x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2)`
Simplifying the expression under the square root:
`x = (5 ± sqrt(25 + 24)) / 4`
`x = (5 ± sqrt(49)) / 4`
`x = (5 ± 7) / 4`
So the two solutions are:
`x = 3`/`2` or `x = -1`
Therefore, `2x^2 - 5x - 3 = 0` when `x = 3/2` or `x = -1`.
Thats wrong
I apologize for the error in my previous response. Let me correct my mistake.
To solve `2x^2 - 5x - 3 = 0`, we can use the quadratic formula:
`x = (-b ± sqrt(b^2 - 4ac)) / 2a`
where `a`, `b`, and `c` are the coefficients of the quadratic equation (`ax^2 + bx + c = 0`).
In this case, `a = 2`, `b = -5`, and `c = -3`, so we substitute these values into the formula:
`x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2)`
Simplifying the expression under the square root:
`x = (5 ± sqrt(49)) / 4`
`x = (5 ± 7) / 4`
So the two solutions are:
`x = 3` or `x = -1/2`
Therefore, `2x^2 - 5x - 3 = 0` when `x = 3` or `x = -1/2`.
Thats correct
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Aubrey says the DOMAIN of the relationship shown here is `0\le x\le1`, but that's not correct.
Edit the numbers in the inequality below to correct her mistake
0\le x\le1
Aubrey's mistake is that she mistakenly assumed that the domain of the given relationship is the same as the possible values of `x` in the inequality `0 ≤ x ≤ 1`. However, this is not necessarily true and there could be other values of `x` that satisfy the given relationship.
To correct her mistake, we need to provide the correct domain of the relationship without making any assumptions based on the inequality `0 ≤ x ≤ 1`. Depending on the specific relationship given, the domain could be different.
If you could provide me with the given relationship, I can help you determine the correct domain.
the relationship is (1,-1) and (0,-2.25)
The two given points `(1, -1)` and `(0, -2.25)` represent two specific values of `x` and `y` that satisfy the relationship. However, these two points alone do not give us enough information to determine the domain of the relationship.
Do you have any additional information or context for the given relationship? For example, is it defined algebraically, graphically, or verbally? Or are there any restrictions or assumptions about the domain that we should be aware of?
Without more information, we cannot determine the correct domain of the relationship.
0\le x\le1
The inequality `0 ≤ x ≤ 1` is a range of possible values of `x` that satisfies the inequality. However, this does not necessarily represent the entire domain of the relationship.
Based on the two given points `(1, -1)` and `(0, -2.25)`, we can construct a linear relationship of the form `y = mx + b` where `m` is the slope and `b` is the y-intercept. We can use these two points to solve for `m` and `b`:
`m = (y2 - y1) / (x2 - x1) = (-2.25 - (-1)) / (0 - 1) = -1.25`
`b = y - mx = -2.25 - (-1.25) * 0 = -2.25`
Therefore, the relationship can be written as `y = -1.25x - 2.25`. The domain of this linear relationship is all real numbers, because we can substitute any real number for `x`. Therefore, the correct domain of the relationship is `-∞ < x < ∞`.
Given `g\left(x\right)=x^{2}-3x-10`, what are the zeros and why
To find the zeros of `g(x) = x^2 - 3x - 10`, we need to solve the equation `g(x) = 0`. We can do this by factoring or by using the quadratic formula.
If we try factoring, we need to find two numbers that multiply to `-10` and add up to `-3`. These numbers are `2` and `-5`, so we can rewrite the equation as:
`g(x) = (x + 2)(x - 5) = 0`
Now we can set each factor equal to zero and solve for `x`:
`x + 2 = 0`; `x - 5 = 0`
`x = -2`; `x = 5`
Therefore, the zeros of `g(x) = x^2 - 3x - 10` are `-2` and `5`. These are the values of `x` where the graph of the function intersects the x-axis. Alternatively, we could have used the quadratic formula to solve for the zeros:
`x = (-b ± sqrt(b^2 - 4ac)) / 2a`
where `a = 1`, `b = -3`, and `c = -10`. Substituting these values, we get:
`x = (3 ± sqrt(3^2 - 4(1)(-10))) / 2(1)`
`x = (3 ± sqrt(49)) / 2`
`x = (3 ± 7) / 2`
`x = -2`; `x = 5`
which are the same zeros we found by factoring.
What are the factors for that last question?
Sure! To factor `g(x) = x^2 - 3x - 10`, we need to find two numbers whose product is `-10` and whose sum is `-3`. These numbers are `-2` and `5`, so we can write:
`g(x) = (x + 2)(x - 5)`
Now we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero, to solve for the zeros of `g(x)`:
`(x + 2)(x - 5) = 0`
Either `x + 2 = 0` or `x - 5 = 0`.
If `x + 2 = 0`, then `x = -2`.
If `x - 5 = 0`, then `x = 5`.
Therefore, the zeros of `g(x) = x^2 - 3x - 10` are `-2` and `5`, and the factored form of `g(x)` is `(x + 2)(x - 5)`.
A pizza maker determined an annual profit in dollars from selling pizzas using `f\left(n\right)=65n-0.04n^{2}`where `n` is the number of pizzas sold. What is the annual profit if the pizza maker sells 300 pizzas?
To find the profits from selling 300 pizzas, we can simply substitute `n = 300` into the equation for the profit function:
`f(n) = 65n - 0.04n^2`
`f(300) = 65(300) - 0.04(300)^2`
`f(300) = 19500 - 0.04(90000)`
`f(300) = 19500 - 3600`
`f(300) = 15900`
Therefore, if the pizza maker sells 300 pizzas, the annual profit would be `$15,900`.
Given `f\left(x\right)=\frac{1}{2}\left(3-x\right)^{2},`
what is the value of `f(15)`?
To find `f(15)`, we need to substitute `x = 15` into the equation for `f(x)`:
`f(x) = 1/2 (3 - x)^2`
`f(15) = 1/2 (3 - 15)^2`
`f(15) = 1/2 (-12)^2`
`f(15) = 1/2 (144)`
`f(15) = 72`
Therefore, `f(15) = 72`.
Suppose you are a solving a quadratic equation using the quadratic formula and this is your work so far:
2x2+7x−15=0
x=
−7 ±72−42−15
22
x=
−7 ±???
4👈
What number goes in place of ???
We have `x = (-7 ± sqrt(7^2 - 4(2)(-15))) / 2(2)`
Simplifying the expression under the square root, we get:
`x = (-7 ± sqrt(49 + 120)) / 4`
`x = (-7 ± sqrt(169)) / 4`
`x = (-7 ± 13) / 4`
So the two solutions are:
`x = -5/2` or `x = 3`
Therefore, the number that goes in place of `???` is `169`.
Find the values of a, b, and c for the quadratic equation below (remember `ax^{2}+bx+c=0`)
`x^{2}-7x=4`
To find the values of `a`, `b`, and `c` for the quadratic equation `ax^2 + bx + c = 0` given `x^2 - 7x = 4`, we need to rewrite the equation in standard form by moving the constant term to the right-hand side:
`x^2 - 7x - 4 = 0`
Now we can see that `a = 1`, `b = -7`, and `c = -4`.
Therefore, the values of `a`, `b`, and `c` for the quadratic equation `ax^2 + bx + c = 0` given `x^2 - 7x = 4` are:
`a = 1`, `b = -7`, and `c = -4`.
If 18 students occupy `3/5`of the seats in the classroom, how many students would occupy `2/3` of the seats in the room?
If 18 students occupy `3/5` of the seats in the classroom, we can set up a proportion to find how many students would occupy `2/3` of the seats:
`(number of students)/(total number of seats) = (number of students occupying 3/5 of the seats) / (total number of seats occupied)`
Let `x` be the total number of seats in the classroom. Then we know that `18` occupy `3/5