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3241004551 [841]
3 years ago
12

Is it possible to form a triangle with the side lengths of 3, 4, 6 ?

Mathematics
2 answers:
pishuonlain [190]3 years ago
8 0
Yes it is possible because there are right triangles, acute, isosceles and way more. So yes, you can make a triangle.
kow [346]3 years ago
7 0

Answer:

yes, it is

Step-by-step explanation:

it will be an obtuse scalene triangle.

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How do I do number 2 ? Someone please helpp w
Alexxandr [17]

Answer:

yes it is a function

Step-by-step explanation:

You have to take the numbers and see if any of them are the same in the X column if their are some of the same numbers you only write the number once. than you connect the numbers to the given output. If any numbers have 2 outputs it is not a function.

7 0
3 years ago
(2x +1) -7 (-6x+9) what is the answer if using the Distributive Property?
Margarita [4]

Answer:

Step-by-step explanation:

Distributive property:  a(b +c) = a*b + a*c

(2x + 1) - 7(-6x + 9) = (2x + 1) + (-7)*(-6x) + (-7)*9

                              = 2x + 1 + 42x - 63  {Combine like terms}

                              = 2x + 42x + 1 - 63

                              = 44x - 62

5 0
3 years ago
Find a general solution of y" + 4y = 0.
klio [65]

Answer:

y(x)=C_{1}cos2x+C_{2}sin(2x)

Step-by-step explanation:

It is a linear homogeneous differential equation with constant coefficients:

y" + 4y = 0

Its characteristic equation:

r^2+4=0

r1=2i

r2=-2i

We use these roots in order to find the general solution:

y(x)=C_{1}cos2x+C_{2}sin(2x)

8 0
3 years ago
Line Segment DE is parallel to side BC of right triangle ABC. CD = 3, DE = 6, and EB = 4. Compute the area of quadrilateral BCDE
dybincka [34]

The area of quadrilateral BCDE = 20.4 sq. units

Let AD = x and AE = y.

Since ΔABC and ΔAED are similar right angled triangles, we have that

AC/AD = AB/AE

AC = AD + CD

= x + 3.

Also, AB = AE + EB

= y + 4

So, AC/AD = AB/AE

(x + 3)/x = (y + 4)/y

Cross-multiplying, we have

y(x + 3) = x(y + 4)

Expanding the brackets, we have

xy + 3y = xy + 4x

3y = 4x

y = 4x/3

In ΔAED, AD² + AE² = DE².

So, x² + y² = 6²

Substituting y = 4x/3 into the equation, we have

x² + y² = 6²

x² + (4x/3)² = 6²

x² + 16x²/9 = 36

(9x² + 16x²)/9 = 36

25x²/9 = 36

Multiplying both sides by 9/25, we have

x² = 36 × 9/25

Taking square root of both sides, we have

x = √(36 × 9/25)

x = 6 × 3/5

x = 18/5

x = 3.6

Since y = 4x/3,

Substituting x into the equation, we have

y = 4 × 3.6/3

y = 4.8

To find the area of quadrilateral BCDE, we subtract the area of ΔAED from area of ΔABC.

So, area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of ΔABC = 1/2 AC × AB

= 1/2 (x + 3)(y + 4)

= 1/2(3.6 + 3)(4.8 + 4)

= 1/2 × (6.6)(8.8)

= 1/2 × 58.08

= 29.04  square units

area of ΔAED = 1/2 AD × AE

= 1/2xy

= 1/2 × 3.6 × 4.8

= 1/2 × 17.28

= 8.64 square units

area of quadrilateral BCDE = area of ΔABC - area of ΔAED

area of quadrilateral BCDE = 29.04 sq units - 8.64 sq units

area of quadrilateral BCDE = 20.4 sq. units

So, the area of quadrilateral BCDE = 20.4 sq. units

Learn more about area of a quadrilateral here:

brainly.com/question/19678935

4 0
2 years ago
Find the sale price Original price: $50 Discount: 15% Sale price: $​
Tema [17]
I hope that i helped ;D

8 0
2 years ago
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