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elena-s [515]
3 years ago
8

How would you begin to plot the ordered pair ( 6,2)?

Mathematics
2 answers:
ryzh [129]3 years ago
5 0

Answer:

Make a grid and put your order pair in the graph

because you would take 6,2 and on the grid go up to 6 and down all the to 2 i hope i helped!

Step-by-step explanation:

Simora [160]3 years ago
3 0

Step-by-step explanation:

start with x which is 6

then plot it

then do y which is 2

and plot

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Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure
Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}

Now, we need to express the quadratic polynomials using their roots, as follows:

ay^2+by+c=a(y-y_1)(y-y_2)

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

8 0
1 year ago
What does a equal in 1/2(a-7)=33?​
Dmitriy789 [7]

To find a, you need to isolate/get the variable "a" by itself in the equation:

\frac{1}{2} (a-7)=33    Multiply the inverse of 1/2, which is 2, to get rid of the fraction

(2)\frac{1}{2}(a-7)=(2)33

a - 7 = 66      Add 7 on both sides to get "a" by itself

a - 7 + 7 = 66 + 7

a = 73

6 0
3 years ago
At a bake sale, as student spent $11.00 buying 3 brownies and 5 cookies. His friend spent $3.95 buying 1 brownie and 2 cookies.
riadik2000 [5.3K]

Answer:

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3b + 5c = 11

b+2c = 3.95

Times 3 on both sides on the second equation above to get the following:

3b + 6c = 11.85

3b+6c - (3b+5c) = 11.85-11

3b+6c-3b-5c = 0.85

c = 0.85

To find the value of b, use substitution:

b+2c = 3.95

b = 3.95-2c

= 3.95 - 2\times0.85

= 2.25

\therefore the cost of a brownie is $$2.25

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4 0
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MOON The expression gives the weight of an object on the Moon in pounds with a weight of w pounds on Earth. What is the weight o
dybincka [34]

Answer: 29.7 pounds

Step-by-step explanation:

The expression in question is w/6.

If the weight of an object on earth is w, the weight of the object on the moon is w/2.

The weight of the space suit is therefore:

= 178.2/6

= 29.7 pounds

4 0
3 years ago
Look at this graph.
Phoenix [80]

Answer:  y= 1/4 x + 500

Step-by-step explanation:

The slope is the difference in y-values divided by difference in x-values plus the starting constant value (500 on the graph)

700 - 600  is the difference in y-values = 100

800 - 400 is the difference in x-values  = 400

100/400 = 1/4

y = x/4 +500

7 0
3 years ago
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