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motikmotik
3 years ago
11

What type of information do you need to prove that two triangles are congruent?

Mathematics
2 answers:
bekas [8.4K]3 years ago
7 0

I need to learn about my solcoil studdyes



Stels [109]3 years ago
3 0
To prove that two triangles are congruent you need to find their perimeter and if they equal the same thing they are equal.
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Ashley paid $14.46 for a pack of 10 pairs of sport socks. What was the price of a pair of sport socks?
Burka [1]
144.6
$14.46x10=
144.6
6 0
3 years ago
An architect is creating a scale drawing of a school computer lab. The length of the lab is 32 feet and the width of the lab is
just olya [345]
The scale drawing of the computer lab will have a length of 4 centimeter and a widtgh of 6 centimeter.
3 0
3 years ago
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What is the measure of Arc ED in circle H? 98 POINTS!!
svlad2 [7]

Answer:

90°

Step-by-step explanation:

EG is a straight line

EG = EHD + DHG

DHG is a 90 degree angle

180 = EHD +90

Subtract 90 from each side

180-90= EHD +90-90

90 = EHD

The arc has the same measure as the angle since it is from the center

arc ED = 90

3 0
3 years ago
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Does anyone know the minim value of the function f(x) =2x^2 -4x -6?. I’m online due to my quarantine and am looking for extra he
navik [9.2K]

Answer:

f(x) = - 8

Explanation:

The given function is

f(x) =2x^2 -4x -6

The first step is to find the derivative of the function. Recall, if

y = ax^b

y' = abx^(b - 1)

Thus,

f'(x) = 4x - 4

We would equate f'(x) to zero and solve for x. We have

4x - 4 = 0

4x = 4

x = 4/4

x = 1

We would substitute x = 1 into the original function and solve for f(x) or y. It becomes

f(1) =2(1)^2 -4(1) - 6 = 2 - 4 - 6

f(1) = - 8

Thus, the minimum value is f(x) = - 8

4 0
10 months ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
3 years ago
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