Write an equation of the line that passes through the given points. (-2,5) and (1,2)

1 answer:

7 0

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-3}{1+2}\implies \cfrac{-3}{3}\implies -1$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-1}[x-\stackrel{x_1}{(-2)}] \\\\\\ y-5=-(x+2)\implies y-5=-x-2\implies y=-x+3$

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Which inequality is true use the number line to help

Nata [24]

Answer:

-3 1/2 > -4.5

Step-by-step explanation:

1.5 > 2 1/2 False

1/2 > 0.5 False

-2.5 > -1.5 False

-3 1/2 > -4.5 True

1.5 > 2 1/2

1/2 = 0.5

2 + 0.5 = 2.5

1.5 is not > 2.5

1.5 < 2.5

1/2 > 0.5

1/2 = 0.5

0.5 is not > 0.5

0.5 = 0.5

-2.5 > -1.5

Negative numbers are oposite of positive numbers when using inequalities. Just as 0 < 1, -1 < 0.

-2.5 is not > -1.5

-2.5 < -1.5

-3 1/2 > -4.5

1/2 = 0.5

3 + 0.5 = 3.5

-3.5 > -4.5

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3 years ago

Read 2 more answers

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

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3 years ago

For quadrilateral ABCD, determine the most precise name for it. A(-2,3), B(9,3), C(5,6) and D(2,6). Show your work and explain.

horsena [70]

To solve this problem you must apply the proccedure shown below:

1. You have the following points given in the problem above:

A<span>(-2,3), B(9,3), C(5,6) and D(2,6)

2. When you plot them, you obtain the figure shown in the graph attached.

3. Therefore, as you can see, the answer is: the figure is a trapezoid, which is define as a quadrilateral with two parallel sides.</span>