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lidiya [134]
3 years ago
15

A=1/2h(a+b) solve for b

Mathematics
1 answer:
vovangra [49]3 years ago
3 0

Answer:

b=2A/h-a

Step-by-step explanation:

A=1/2h(a+b)

a+b=A/(1/2h)

a+b=A/(h/2)

a+b=(A/1)(2/h)

a+b=2A/h

b=2A/h-a

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1. Given the equation (x – 3)2 – 32 = 17 a)
m_a_m_a [10]
We know that

The square root property<span> is one method that is used to find the solutions to a quadratic equation.  This method involves taking the square roots of both sides of the equation.  
</span>(x – 3)² – 32 = 17-------> (x – 3)² = 17+32-----------> (x – 3)² = 49
(x – 3)² = 7²--------> <span>will take the square root of each side
</span>so
√(x – 3)²=(+-)√7²------> <span>notice the use of the </span> sign (+-), this will give us both a positive and<span>  a negative root
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then

x-3=7--------> x=10
x-3=-7-------> x=-4

the answer Part a) is
x=10
x=-4

Part b) Explain why the given equation has two solutions.

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Because exist<span> the possibility of two roots for every square root, one positive and one negative</span>
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3 years ago
What does y = 2.5x - 10 equal?
mariarad [96]

Answer:

Step-by-step explanation:

x= y+10

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1. What must be true about a circle with a circumference of 16 π?
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1. - B) <span>the diameter is 16

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5. - A.) 9.42

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The 2nd, 6th, 8th terms of an A.P. form a G.P. , find the common ratio and the general term of the G.P.​
melisa1 [442]

The terms of an arithmetic progression, can form consecutive terms of a geometric progression.

  • The common ratio is: \mathbf{r = \frac{a + 5d}{a + d}}
  • The general term of the GP is: \mathbf{a_n = (a + d) \times (\frac{a + 5d}{a + d})^{n-1}}

The nth term of an AP is:

\mathbf{T_n = a + (n - 1)d}

So, the <em>2nd, 6th and 8th terms </em>of the AP are:

\mathbf{T_2 = a + d}

\mathbf{T_6 = a + 5d}

\mathbf{T_8 = a + 7d}

The <em>first, second and third terms </em>of the GP would be:

\mathbf{a_1 = a + d}

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The common ratio (r) is calculated as:

\mathbf{r = \frac{a_2}{a_1}}

This gives

\mathbf{r = \frac{a + 5d}{a + d}}

The nth term of a GP is calculated using:

\mathbf{a_n = a_1r^{n-1}}

So, we have:

\mathbf{a_n = (a + d) \times (\frac{a + 5d}{a + d})^{n-1}}

Read more about arithmetic and geometric progressions at:

brainly.com/question/3927222

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