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3 years ago

Will give brainliest answer

Mathematics

2 answers:

geniusboy [140]3 years ago

8 0

Answer:

<h3> Radius = 13 / 2 = 6.5 cm</h3><h3>Diameter = 13 cm</h3>

Explanation

<h3><u>Radius</u></h3>

The straight line is drawn from the centre of a circle to a point on its circumference is called radius of the circle. The radius of a circle is half of its diameter.

<h3><u>Diameter</u></h3>

The chord that passes through the centre of a circle is called diameter of circle. Diameter is also called the largest chord of any circle. The length of diameter of a circle is two times it's radius.

Hope this helps...

Good luck on your assignment...

topjm [15]3 years ago

3 0

Answer:

Radius = 6.5cm

Diameter = 13cm

Step-by-step explanation:

The diameter is given (13)

Radius is half the diameter (13/2=6.5)

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A line intersects the points

allochka39001 [22]

Answer:

The answer would be -5. Hopefully I'm not wrong.

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2 years ago

Draw a picture to show the product of 4:7 x 3. Then write the product

Hoochie [10]

Answer:

¹²/₇

Step-by-step explanation:

Model ⁴/₇ × 3

Assume that you have three pies, each dived into seven slices.

There are only four slices remaining in each pie, that is, there is ⁴/₇ pie in each pie plate

The picture represents ⁴/₇ × 3.

Model the product

Now, transfer slices to get as many filled pie plates as possible.

Count the total slices.

You have 12 slices, and each slice represents ⅐ of a pie.

You have ¹²/₇ pie.

∴ ⁴/₇ × 3 = ¹²/₇

6 0

3 years ago

PLEASE ANSWER ASAP NEED HELP REALLY BAD CALLING OUT TO ALL MATH PROS WHO CAN HELP ME PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

sergeinik [125]

Answer:

<abd is 90

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this isnt that hard im in middle school and doing this

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

X over 3 - 2 = 5x - 2 , having trouble solving for x or x/3 -2 = 5x -2

Len [333]

I got x=0 because I simplified both sides of the equation and then isolated the equation

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3 years ago

Express the complex number in trigonometric form -6 + 6sqrt3i

marusya05 [52]

$\large\begin{array}{l}\\\\ \textsf{This is a complex number in }\mathsf{a+bi}\textsf{ form:}\\\\ \mathsf{z=-6+6\sqrt{3}\,i}\\\\ \textsf{where }\mathsf{a=-6~~and~~b=6\sqrt{3}.} \end{array}$

$\large\begin{array}{l}\\\\ \bullet~~\textsf{1}\mathsf{^{st}}\textsf{ step: Find the absolute value of z (complex modulus of z):}\\\\ \mathsf{|z|=\sqrt{a^2+b^2}}\\\\ \mathsf{|z|=\sqrt{(-6)^2+(6\sqrt{3})^2}}\\\\ \mathsf{|z|=\sqrt{36+36\cdot 3}}\\\\ \mathsf{|z|=\sqrt{36\cdot 108}}\\\\ \mathsf{|z|=\sqrt{144}}\\\\ \mathsf{|z|=12} \end{array}$

$\large\begin{array}{l}\bullet~~\textsf{2}\mathsf{^{nd}}\textsf{ step: Find the argument }\mathsf{\theta}\textsf{ of z:}\\\\ \mathsf{arg(z)=\theta}\\\\ \textsf{is an angle that satisfies the following conditions:}\\\\ \begin{cases} \mathsf{cos\,\theta=\dfrac{a}{|z|}=\dfrac{a}{\sqrt{a^2+b^2}}}\\\\ \mathsf{sin\,\theta=\dfrac{b}{|z|}=\dfrac{b}{\sqrt{a^2+b^2}}} \end{cases}\qquad\quad-\pi$

$\large\begin{array}{l}\textsf{For }\mathsf{z=-6+6\sqrt{3}\,i,}\\\\ \begin{cases} \mathsf{cos\,\theta=\dfrac{-6}{12}=-\,\dfrac{1}{2}}\\\\ \mathsf{sin\,\theta=\dfrac{6\sqrt{3}}{12}=\dfrac{\sqrt{3}}{2}} \end{cases}\\\\\\ \textsf{Since}\\\\ \mathsf{cos\,\theta0,}\\\\ \theta\textsf{ lies in the 2}\mathsf{^{nd}}\textsf{ quadrant.}\\\\\\\textsf{Therefore,}\\\\ \mathsf{\theta=\dfrac{2\pi}{3}~rad~~(120^\circ)} \end{array}$

$\large\begin{array}{l}\bullet~~\textsf{3}\mathsf{^{rd}}\textsf{ step: Express z in the trigonometric form:}\\\\ \begin{array}{rcl} \mathsf{z}&\!\!=\!\!&\mathsf{|z|\cdot (cos\,\theta+i\,sin\,\theta)}\\\\ \mathsf{-6+6\sqrt{3}\,i}&\!\!=\!\!&\mathsf{12\cdot \left(cos\, \dfrac{2\pi}{3}+i\,sin\,\dfrac{2\pi}{3}\right)\qquad\quad\checkmark} \end{array} \end{array}$

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$\large\textsf{I hope it helps.}$

Tags: <em>complex number trigonometric form trig absolute value modulus argument angle</em>

4 0

3 years ago