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ivann1987 [24]
3 years ago
14

NEED HELP ASAP AGAIN

Mathematics
2 answers:
balu736 [363]3 years ago
8 0
4)
A=1/2 bh
A = 1/2(6)(3)
A = 9 in^2 

5)
A=1/2 bh
A=1/2(7)(10)
A = 35 m^2

6)
A=1/2 bh
A=1/2(5)(8)
A=20 ft^2

7)
A=1/2 (b1+b2)h
A=1/2 (5+10)8
A=60 in^2 

8)
A=1/2 (b1+b2)h
A=1/2 (13+19) 10
A=160 mm^2

9)
A=1/2 (b1+b2)h
A=1/2 (3+6) 2
A = 9 ft^2


Reptile [31]3 years ago
7 0
4)A=1/2 bh
A = 1/2(6)(3)
A = 9 in^2 

5)A=1/2 bh
A=1/2(7)(10)
A = 35 m^2

6)A=1/2 bh
A=1/2(5)(8)
A=20 ft^2

7)A=1/2 (b1+b2)h
A=1/2 (5+10)8
A=60 in^2 

8)A=1/2 (b1+b2)h
A=1/2 (13+19) 10
A=160 mm^2

9)A=1/2 (b1+b2)h
A=1/2 (3+6) 2
A = 9 ft^2

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Feliz [49]

Answer:

true

Step-by-step explanation:

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5 0
3 years ago
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Solve for x<br> X^2+6x+9=20
Masja [62]

Answer:

x = 1.47  or x = -7.47

Step-by-step explanation:

x²+6x+9=20

This is a quadratic equation

x²+6x+9-20=0

x²+6x-11=0

Step 1 : Write the quadratic formula

x = <u>-b±√b²-4(a)(c)</u>

              2a

Step 2 : Substitute values in the formula

a = 1

b = 6

c = -11

x = <u>-6±√6²-4(1)(-11)</u>

              2(1)

x = <u>-6±√80</u>

          2

x = -3 + 2√5 or x = -3 - 2√5

x = 1.47         or x = -7.47

!!

4 0
3 years ago
Read 2 more answers
If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?
lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

3x^2-(3k-2)x-(k-6) \text{ with }k=2:

3x^2-(3\cdot 2-2)x-(2-6)

3x^2-4x+4

I'm going to solve 3x^2-4x+4=0 for x using the quadratic formula:

\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}

\frac{4\pm \sqrt{16-16(3)}}{6}

\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}

\frac{4\pm 4\sqrt{-2}}{6}

\frac{2\pm 2\sqrt{-2}}{3}

\frac{2\pm 2i\sqrt{2}}{3}

Let's see if uv=u+v holds.

uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}

Keep in mind you are multiplying conjugates:

uv=\frac{1}{9}(4-4i^2(2))

uv=\frac{1}{9}(4+4(2))

uv=\frac{12}{9}=\frac{4}{3}

Let's see what u+v is now:

u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}

u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}

We have confirmed uv=u+v for k=2.

4 0
3 years ago
What is the equation of a line with slope -2 and passes through the point (0, 3)?
kramer
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5 0
3 years ago
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faltersainse [42]
I think is should be (5,0)
3 0
3 years ago
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