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3 years ago

15

Two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle. Which congrue

nce theorem can be used to prove that the triangles are congruent?
PLEASE ANSWER FAST I HAVE 5 MINUTES LEFT ON THIS TEST

SSS
AAS
SAS
HL

2 answers:

vesna_86 [32]3 years ago

6 0

It is AAS - hope it helps

-BARSIC- [3]3 years ago

5 0

<span>The answer is AAS. </span><span>Angle Angle Side </span>postulate<span> </span><span>states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.</span>

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-6.3x+14 and 1.5x-6<br> answer in simplified form

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Answer:

The simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8

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If it is 375 miles form Los Angeles To San Francisco and on a map the scale is 2cm=10miles how far apart are they on the map

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Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span> $S_{a}=S_{l}-2$
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span> $t_{l}=t_{a}+1$
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span> $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

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