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Kryger [21]
3 years ago
6

The first and last terms of a 10-term arithmetic series are listed in the table. What is the sum of the series?

Mathematics
1 answer:
snow_tiger [21]3 years ago
4 0
The sum of an arithmetic series:
S_n=\frac{n(a_1+a_n)}{2}

a_1=3 \\
a_{10}=75 \\ \\
S_{10}=\frac{10(a_1+a_{10})}{2}=\frac{10(3+75)}{2}=5(3+75)=5 \times 78=390

The sum of the series is 390.
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59

Step-by-step explanation:

alternate interior angles are congruent

6 0
2 years ago
which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the l
victus00 [196]
One equation for this would be

y = \frac{41}{16} x-\frac{55}{8}

We start by finding the slope between the two points:

m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}
\\
\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})
\\
\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})
\\
\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

y-y_1=m(x-x_1)
\\
\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})
\\
\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}
\\
\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}
\\
\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}
\\
\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}
\\
\\y=\frac{41}{16}x-\frac{550}{80}
\\
\\y=\frac{41}{16}x-\frac{55}{8}
6 0
3 years ago
4. You sketch a right triangle that has a
stich3 [128]

Answer:

√95 cm

Step-by-step explanation:

To solve, you need to use the pythagorean theorem, or a^2 + b^2 = c^2

The hypotenuse is the c and let one leg be b. You can write:

a^2 + 7^2 = 12^2

a^2 + 49 = 144

Now, you need to solve for a:

a^2 = 144 - 49

a^2 = 95

a = √95 cm, or about 9.75cm

3 0
3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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