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Whitepunk [10]
3 years ago
13

PROVE THAT TWO TRIANGLES ARE CONGRUENT USING THE SAS CONGRUENCE CRITERIA PLEASEEEEE I NEED ASAP DUE IN 15 MINS PLEEEEASE​

Mathematics
2 answers:
Klio2033 [76]3 years ago
8 0

Answer:

Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

Step-by-step explanation:

zimovet [89]3 years ago
6 0

Answer:

Step-by-step explanation:

The length of DE is congruent to the length of FG, which is sqrt(5). The length of EF and GH are congruent too. Two pairs of sides are congruent, one pair of angles are congruent (vertical), SAS

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Oil spill spreads 25 m² every 1/6 hour how much area does the oil spill cover after two hours
Delvig [45]
1/6 of an hour is 10 minutes because there are 60 minutes in an hour so 60/6=10
so it's 25m^2/10minutes
now just multiply 25 by 12 ( 12 lots of 10 minutes in 2 hours)
25*12=300
answer is 300m^2

hope this helped
6 0
2 years ago
Read 2 more answers
You start at (3, 0). You move down 5 units and right 1 unit. Where do you end?
nevsk [136]

Answer:

Pretty sure it's (4, -5)

Step-by-step explanation:

If not, shoot me I guess.

8 0
3 years ago
Someone please explain how this make sense
Andru [333]

Answer:

6x + 12y = 816

Step-by-step explanation:

6 dollars per regular admission

12 dollars per both dance and volleyball

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3 0
2 years ago
How to do number 2 and 5 !!! Please someone help
Brums [2.3K]
2a) there is a right angle on T so 180-90-37=53
2b) do pythogrean theorem 22sq+ 12sq= 628 sq rt of 628 is 25.06 which is 25
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4 0
3 years ago
Evaluate the integral ∫2−1|x−1|dx
defon

I think you might be referring to the definite integral,

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx

Recall the definition of absolute value:

|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

Then |x-1|=x-1 if x\ge1, and |x-1|=1-x is x. So spliting up the integral at <em>x</em> = 1, we have

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx

The rest is simple:

\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}

5 0
3 years ago
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