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3 years ago

Which circle shows AB that measures 60°? 60° 0 60° A60°

Mathematics

2 answers:

Mrac [35]3 years ago

7 0

Answer: is there a picture to look at?

PtichkaEL [24]3 years ago

6 0

Answer:

The answer is D

Step-by-step explanation:

Have a great day :)

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(-6)(2) + (-8÷4) what is the anwer?

Novay_Z [31]

Answer:

-14

Step-by-step explanation:

(-6)(2)= -12

(-8÷4)= -2

-12+(-2)= -14

8 0

3 years ago

7x + 3x + x⁴ divided by x-2 using long division

VARVARA [1.3K]

Answer:

10x + x^4/x-2

Step-by-step explanation:

Combine like terms

(10x + x^4)/(x-2)

4 0

1 year ago

Read 2 more answers

72 + 8x + 18x = 36x

ArbitrLikvidat [17]

$72 + 8x + 18x = 36x
$

$72+26x=36x
$

$72=36x-26x
$ $\text{Subtract 26x from both sides}$

$72=10x
$

$\text{Divide both sides by 10}$

$x=36/5$

4 0

3 years ago

Simplify the expression and write the result in standard form, a+bi.<br><br> 6 - 8i / -2

gregori [183]

Answer:

$\frac{6 - 8i}{ - 2} = \frac{6}{ - 2} - \frac{8i}{ - 2} (splitting \: denomenator) \\ = - 3 - - 4i \\ = - 3 + 4i \\ thank \: you$

4 0

3 years ago

Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation

lapo4ka [179]

Answer:

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=73$

Step-by-step explanation:

$First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}$

$a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\$

$S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\$

$S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]$

$S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73$

5 0

3 years ago