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Paul [167]
3 years ago
10

Please help me find the area of the triangles in the picture shown below.

Mathematics
2 answers:
LenKa [72]3 years ago
7 0

Answer:

1)6

2)20

3)45

Step-by-step explanation:

a=1/2 bh

1) 4*3=12/2=6

2) 8*5=40/2=20

3)15*6=90/2=45

lyudmila [28]3 years ago
3 0

Answer:

1) 6

2) 20

3) 45

Step-by-step explanation:

To find the area of a triangle, use the formula \frac{1}{2} bh, where b represents the length of the base, and h represents the height. In the first triangle, the base length is 4 and the height is 3, so the area is 6. In the second triangle, the base length is 5 and the height is 8, so the area is 20. In the third triangle, the base length is 15 and the height is 6, so the area is 45.

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-3z+8=2z-12 what is the answer?
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Answer:

Step-by-step explanation:

Simplifying

-3a + 8 = 2z + -12

Reorder the terms:

8 + -3a = 2z + -12

Reorder the terms:

8 + -3a = -12 + 2z

Solving

8 + -3a = -12 + 2z

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-8' to each side of the equation.

8 + -8 + -3a = -12 + -8 + 2z

Combine like terms: 8 + -8 = 0

0 + -3a = -12 + -8 + 2z

-3a = -12 + -8 + 2z

Combine like terms: -12 + -8 = -20

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a = 6.666666667 + -0.6666666667z

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3 years ago
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Find the radius of convergence, then determine the interval of convergence
galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

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  • If β > 1, then the series diverges
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Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

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We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

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