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2 years ago

Can you please solve this

Mathematics

1 answer:

ddd [48]2 years ago

7 0

The answer to your question is a=124 because the center triangle is a isosceles triangle

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20 POINTS! Please help, best answer will recessive brainliest answer points as well as 5 stars and a thank you. :)

valentina_108 [34]

Answer:

(3) Diagonals are perpendicular

Step-by-step explanation:

5 0

2 years ago

In 1985, there were 285 cell phone subscribers in Arlington. The number of subscribers increased by 75% per year since 1985. Fin

ANTONII [103]

Answer:

The number of subscribers in 2008 is 110,845,988

Step-by-step explanation:

The number of cell phone subscribers in Arlington, in t years after 1985, can be modeled by the following equation.

$N(t) = N(0)(1+r)^{t}$

And which N(0) is the number of subscribers in 1985 and r is the rate it increases per year, as a decimal.

In 1985, there were 285 cell phone subscribers in Arlington.

This means that $N(0) = 285$

The number of subscribers increased by 75% per year since 1985.

This means that $r = 0.75$

$N(t) = 285(1.75)^{t}$

Find the number of subscribers in 2008.

2008 is 2008-1985 = 23 years after 1985. So we have to find N(23).

$N(23) = 285(1.75)^{23} = 110,845,988$

The number of subscribers in 2008 is 110,845,988

8 0

2 years ago

What is the measure for angle CBD <br> PLSSS HELPP!!!!

Andrei [34K]

The measure of the angle CBD = 180° - 30° = 150°.

3 0

2 years ago

A cylinder has a base area of 44.6 mm2 and a height of 24 mm.

Free_Kalibri [48]

Volume of cylinder = base area x height
= 44.6 x 24
= 1070.4 mm³

7 0

3 years ago

Read 2 more answers

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. =)

6 0

3 years ago

Read 2 more answers

Can you please solve this​

Can you please solve this