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Sliva [168]
2 years ago
8

Which of the following coordinates belong in the box

Mathematics
1 answer:
Yuri [45]2 years ago
8 0
I need a picture to answer your question
You might be interested in
Graph ABCD if A(1, 4), B(6, 6), C(4, 1), and D(−1, −1). Is ABCD a rhombus? Show how you know.
stich3 [128]

C4,1 because I solved it and it showed me

6 0
3 years ago
QUESTION TWO Determine whether or not a constant function can be
Sveta_85 [38]
<h3>Answers:</h3>
  • (a) It is <u>never</u> one-to-one
  • (b) It is <u>never</u> onto

=========================================================

Explanation:

The graph of any constant function is a horizontal flat line. The output is the same regardless of whatever input you select. Recall that a one-to-one function must pass the horizontal line test. Horizontal lines themselves fail this test. So this is sufficient to show we don't have a one-to-one function here.

Put another way: Let f(x) be a constant function. Let's say its output is 5. So f(x) = 5 no matter what you pick for x. We can then show that f(a) = f(b) = 5 where a,b are different values. This criteria is enough to show that f(x) is not one-to-one. A one-to-one function must have f(a) = f(b) lead directly to a = b. We cannot have a,b as different values.

----------------------------

The term "onto" in math, specifically when it concerns functions, refers to the idea of the entire range being accessible. If the range is the set of all real numbers, then we consider the function be onto. There's a bit more nuance, but this is the general idea.

With constant functions, we can only reach one output value (in that example above, it was the output 5). We fall very short of the goal of reaching all real numbers in the range. Therefore, this constant function and any constant function can never be onto.

4 0
2 years ago
in order to solve the system of equations below , harvey multiplies each equation by a constant to eliminate the x terms.
kicyunya [14]

Answer:

14x + 6y = 10 \\  - 14x - 35y = 77

multiply all terms in the top equation by 2

multiply all terms in the second equation by -7. it would then add the two equations together to cancel out the variable x

You will then have two equations in terms of y

solve for y and either equation then substitute that value for y into the other equation then solve for x

6 0
3 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
zloy xaker [14]

Answer:

t=-20ln\left(\dfrac{1}{3}\right)

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

A=45e^{-0.05t}

We want to determine the value of t for which the area, A(t)=15 square kilometers.

15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0
3 years ago
Solve for w<br> 5 (w - 1) - 6 = -2 (-6w+7) -w<br> Simplify your answer as much as possible.
Soloha48 [4]
It would be w=3/6 or 1/2
8 0
3 years ago
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