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julsineya [31]
3 years ago
15

How many different triangles can you make if you are given the measurements for two angles and the length of a side that is NOT

shared?
A.0

B. 1

C. 3

D. 2

E.many
Mathematics
1 answer:
shepuryov [24]3 years ago
7 0
I believe it’s only one triangle because if you have two angles already the third can’t be switched
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Mrs. Robert’s vegetable garden is 6 feet wide and 8 feet long. A drawing of the garden has a scale of 1 inch 2 feet. What is the
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Step-by-step explanation:

It’s 24

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What is the answer for this question ??
WITCHER [35]

Answer:

1 236^2

Step-by-step explanation:

first find the area of the square field .

Side × side

100m×100m=10 000m^2

secondly find the area of the whole shape .

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side ×side

106m×106m=11 236m^2

lastly you subtract to get area of the path

11 236-10 000=1 236m^2

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Solve each system by substitution<br>x - y=4<br>x+2y=-2​
DENIUS [597]

Answer:

<h2>x = 2 and y = -2</h2>

Step-by-step explanation:

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3 years ago
A quadratic equation is shown below: 3x2 − 15x + 20 = 0 Part A: Describe the solution(s) to the equation by just determining the
Triss [41]

Answer:

\text{The roots of }3x^2+5x-8=0\text{ are }x=1,\frac{-8}{3}

Step-by-step explanation:

\text{Part A: Given a quadratic equation }3x^2-15x+20=0  

\text{Comparing above equation with }ax^2+bx+c=0  

a=3, b=-15, c=20

Discriminant can be calculated as

D=b^2-4ac

D=(-15)^2-4(3)(20)=225-240=-15

The roots are imaginary

The solution is

x=\frac{-b\pm\sqrt{D}}{2a}

x=\frac{-(-15)\pm \sqrt{-15}}{2(3)}=\frac{15\pm\sqrt{15}i}{6}

The roots are not real i.e these are imaginary    

\text{Part B: Given a quadratic equation }3x^2+5x-8=0  

\text{Comparing above equation with }ax^2+bx+c=0  

a=3, b=5, c=-8

Discriminant can be calculated as

D=b^2-4ac

D=(5)^2-4(3)(-8)=25+96=121>0

The roots are real

By quadratic formula method

The solution is

x=\frac{-b\pm\sqrt{D}}{2a}

x=\frac{-5)\pm \sqrt{121}}{2(3)}=\frac{-5\pm 11}{6}

x=1,\frac{-8}{3}

which are required roots.

I choose this method because I can get the solutions directly by substituting the values in formula, and I don't have to guess the possible solutions.

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