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IrinaVladis [17]
3 years ago
5

In circle E shown, chords AB and CD are parallel. (a) What does this information tell you about 1  and 2  ? Why? (b) Given you

r answer to (a), what can now be said about  AC and  BD? Why?
Mathematics
1 answer:
sp2606 [1]3 years ago
4 0

Answer:

6835

Step-by-step explanation:

58gfbhfm

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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
Which graph shows the solution to the equation below? log Subscript 3 Baseline (x + 3) = log Subscript 0.3 (x minus 1) On a coor
Effectus [21]

Answer:

On a coordinate plane, 2 curves intersect at (1, 1). One curve curves up and to the right from quadrant 3 into quadrant 1. The other curve curves down from quadrant 1 into quadrant 4

Step-by-step explanation:

The first function is given as:

log_3(x+3)

The second function is given as:

log_{0.3}(x-1)

First we graph both the functions.

We can see that one curves up and to the right from quadrant 3 into quadrant 1. This curve is of  log_3(x+3)

The other curve curves down from quadrant 1 into quadrant 3

Both curves interest almost at (1,1)

See the graph attached below

Blue line represents first function

Green line represents second function

The solution lies on the Red line.

8 0
3 years ago
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Work out 6/5 of 200 and show working out
katrin [286]

Answer:

240.

Step-by-step explanation:

200 * 6/5

= 1200/5

= 240

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3 years ago
Mr. Yueng graded his students’ math quizzes.
zubka84 [21]

Answer:

7.3*

Step-by-step explanation:

do the inverse:

22/3=7.3*

*=recurring

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Helpp!!!!!!!!!!!!!!!!!!
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I think it’s X+12=-15
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