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Paul [167]
2 years ago
9

Need help ASAP ! Questions are in the pic

Mathematics
1 answer:
Vedmedyk [2.9K]2 years ago
3 0

Answer: b (-2,3)

Step-by-step explanation:

you substitute the x and y values in (-2,-3) for the x and y in the problem.

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horrorfan [7]
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hope this helped :)
5 0
3 years ago
2. Check the boxes for the following sets that are closed under the given
son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Addition

   c) AdditionSum

   d) in this case we have two possibilities

       * If we move to the right the addition

       * If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

        -3 -1 = -4

        -4 -1 = -5

        -7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) 0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}  ...

In this case we can see more clearly the sequence when writing in this way

      0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}

each term is found by adding 1 to the current term,

      \sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}

Therefore the mathematical operation is the addition

c)   ... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...

      The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

  • To move to the right add 1

           -\frac{-10}{2} + 1 = \frac{-10}{2}  -   \frac{2}{2}  = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}

  • To move left subtract 1

         \frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}

         

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Sum

   c) Sum

   d) This case we have two possibilities

  •  If we move to the right the sum
  •  If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0
2 years ago
MR.Trek made some trail mix for his backpacking team. He used 2 1/2 cups of granola, 1 3/4 cups of peanuts, 1 1/3 cups of raisin
valentina_108 [34]
6 7/12 of trail mix . This is the answer

if you want me to explain then just comment back
6 0
3 years ago
Find the amount financed and the finance charge in dollars for the following problem. Choose the correct answers.
diamong [38]

$350 * 30/100 = $ 105 which is down payment.

$350 - $105 = $245 which is left.

But he will pay $24.50 * 12 = $294 instead.

$294 - $245 = $49

He will pay $49 more.

He paid $105 as down payment and will pay $294 more with 12 months.

Total will be $105 + $294 = $399

4 0
3 years ago
The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
a_sh-v [17]
Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
\\\\\\
BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
\\\\\\
BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
\\\\\\
4[15]\implies 60

7 0
2 years ago
Read 2 more answers
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