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3 years ago

6

jane is practicing solving systems of equations by inspection.she thinks that y=5/6x+2 and y=5/6x-2 would have exactly one solut

ion.is she correct? why or why not?

1 answer:

luda_lava [24]3 years ago

5 0

It's no solution since the lines are parallel to each other. therefore, jane is WRONG
Parallel lines: same slope but different y intercept

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Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

so

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

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3 years ago

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ra1l [238]

Answer:

A. (9,-5)

Step-by-step explanation:

Translation down is -2 to the y coordinate, which is (5,-5). Then you translate it right 4 units, which is +4 to the x coordinate, so the final coordinate is (9,-5).

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Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent

Reptile [31]

Answer:

0.4

Step-by-step explanation:

Given

60 % wear neither ring nor a necklace

20 % wear a ring

30 % wear necklace

This question can be Solved by using Venn diagram

If one person is choosen randomly among the given student the probability that this student is wearing a ring or necklace is

$P\left ( wear \ ring\ or \ necklace )+P\left ( neither\ ring\ or\ necklace)=1$

$P\left ( wear \ ring\ or\ necklace )=1-0.6=0.4$

The sum of probabilty is equal to 1 because it completes the set

Therefore the required probabilty is 0.4

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Travis and Wendy were in an election for club president. Of the club members, 1/3 voted for Travis, 5/12 voted for Wendy, and th

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The answer would be C. 3/12 (Reduced 1/4) (Decimal 0.25) I hope this helped ^^

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