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3 years ago

Question

Mathematics

1 answer:

ad-work [718]3 years ago

7 0

Answer:

Step-by-step explanation:

imagine you chop off a right triangle from the left side

now you have a rectangle and a right triangle

the rectangle

height 5

width 12

5 x 12 = 60

triangle

height 5

width 3

3 x 5 = 15

15 / 2 = 7.5

60 + 7.5 = 67.5 m^2

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A rectangle's length is twice its width. if the perimeter is 66 cm find the width of the rectangle

Lesechka [4]

Perimeter = 2L + 2W

Let x = width and let 2x = length

Perimeter = 66

2x + 2(2x) = 66

2x + 4x = 66 --> 6x = 66 --> x = 66/6 --> x = 11

The width of the rectangle is 11 cm.

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3 years ago

3x/8-1=5/8<br> How do I start to find out the answer to this equation?

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(3x) •(8) =(8-1) •(5) multiply the numerator of the 1st fraction by the denominator of the second fraction.

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Line M contains the following two points: (1, 10) and (6, 20). What is the slope of line M?

iren2701 [21]

Answer:

The slope is 2.

Step-by-step explanation:

In order to find the slope of the line, we need to start by using the points in the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (20 - 10)/(6 - 1)

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So the slope of this line would be 2

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3 years ago

Find the multiplicative inverse of 6 + 2i

Marina CMI [18]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2774989

________________

Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________

The inverse multiplicative of $\mathsf{z=a+bi}$ is

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\
=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}$

$=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\
\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}$

________

For this question,

$\mathsf{z=6+2i}$

So,

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{6+2i}}\\\\\\
=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\
=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\
=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$

$\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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Write the missing value above the variable. M-9=7

tia_tia [17]

M=16
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