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fgiga [73]
3 years ago
5

3x to the second power + 7x + 2

Mathematics
2 answers:
Mrrafil [7]3 years ago
8 0

you cant add 3x^2 and 7x + 2 :)

you put it in the correct order by the degree :)

Andre45 [30]3 years ago
4 0

Answer: 16x+2

1. 3x Squared is going to be 9X + 7x =16x+2

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How many unique triangles can be drawn with side lengths 8 in., 12 in., and 24 in.? Explain. A. Any three sides form a unique tr
netineya [11]

Answer: C. Two of the side lengths add to a sum that is less than the third side length, so these lengths cannot be used to draw any triangles.

Explanation:

Those two sides in question are 8 and 12. They add to 8+12 = 20, but this sum is less than the third side 24. A triangle cannot be formed.

Try it out yourself. Cut out slips of paper that are 8 units, 12 units, and 24 units respectively. The units could be in inches or cm or mm based on your preference.

Then try to form a triangle with those side lengths. You'll find that it's not possible. If we had the 24 unit side be the horizontal base, so to speak, then we could attach the 8 and 12 unit lengths on either side of this horizontal piece. But then no matter how we rotate those smaller sides, they won't meet up to form the third point for the triangle. The sides are simply too short. Other possible configurations won't work either.

As a rule, the sum of any two sides of a triangle must be larger than the third side. This is the triangle inequality theorem.

That theorem says that the following three inequalities must all be true for a triangle to be possible.

  • x+y > z
  • x+z > y
  • y+z > x

where x,y,z are the sides of the triangle. They are placeholders for positive real numbers.

So because 8+12 > 24 is a false statement, this means that a triangle is not possible for these given side lengths. Therefore, 0 triangles can be formed.

6 0
2 years ago
What is the smallest fredholl number that is prime
kaheart [24]
Smallest Fredholl no. that's prime is 13
7 0
3 years ago
Read 2 more answers
identify the equation that does not belong with the other three. explain your reasoning. 6+x=9 | 15= x+12 | x+9 =11 | 7+x=10
Katena32 [7]
Answer is x+9=11

6+x=9
x=3

15=x+12
x=3

7+x=10
x=3

x+9=11
x=2

So x+9=11 has a different x value than the rest of the equations. So it doesn't belong with the other three.
6 0
3 years ago
4x² = 32 What does x equal
babymother [125]

So here is what you would do:

lets just pretend that there is no x in the equation

so this is what the equation would look like:

first, is 4 to the power of 2=16 and now we are going to put the x back into the equation so: 4x to the power of 2, but we did the exponents so, now x is the missing number. We can use the 32 because it is the solution to the equation. So we would do 32-16=16 so the missing number is 16 so x=16

Answer: x=16

4 0
2 years ago
Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE
inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to 20\ units

Part 3) The area is equal to 25\ units^{2}

Step-by-step explanation:

we have

A(5,0), B(2,4), C(-2,1),D(1,-3)

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Find the distance AB

A(5,0),B(2,4)

substitute in the formula

d=\sqrt{(4-0)^{2}+(2-5)^{2}}

d=\sqrt{(4)^{2}+(-3)^{2}}

d=\sqrt{25}

AB=5\ units

Find the distance BC

B(2,4), C(-2,1)

substitute in the formula

d=\sqrt{(1-4)^{2}+(-2-2)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}

BC=5\ units

Find the distance CD

C(-2,1),D(1,-3)

substitute in the formula

d=\sqrt{(-3-1)^{2}+(1+2)^{2}}

d=\sqrt{(-4)^{2}+(3)^{2}}

d=\sqrt{25}

CD=5\ units

Find the distance AD

A(5,0),D(1,-3)

substitute in the formula

d=\sqrt{(-3-0)^{2}+(1-5)^{2}}

d=\sqrt{(-3)^{2}+(-4)^{2}}

d=\sqrt{25}        

AD=5\ units

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

B(2,4),D(1,-3)

substitute in the formula

d=\sqrt{(-3-4)^{2}+(1-2)^{2}}

d=\sqrt{(-7)^{2}+(-1)^{2}}

BD=\sqrt{50}\ units        

<em>Verify if the polygon is a square</em>

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

BD^{2}=AD^{2}+AB^{2}

substitute

(\sqrt{50})^{2}=5^{2}+5^{2}

50=50 -----> is true

so

The triangle BDA is a right triangle

therefore

The polygon is a square

<em>Find the Area of the polygon</em>

The area of a square is equal to

A=b^{2}

we have

b=5\ units

A=5^{2}=25\ units^{2}

<em>Find the perimeter of the polygon</em>

The perimeter of a square is equal to

P=4b

we have

b=5\ units

P=4(5)=20\ units

6 0
2 years ago
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