A student jog the same number of laps each week.9 students jogged 63 laps.How many laps will 13 students jog

Plz help I am giving away 20 points!!!!!!!!

vova2212 [387]

Okay, so on number one I have no idea. I keep getting 20 because 14 - (-3)2 = 14- (-6) which is really 14 + 6=20.

Number 2 is r+15=61

Number 3 I didn't get any of those answers because you have to solve what is in the parenthesis first so3*5= 15+1=16 then you multiply that by two giving you 16*2=32 and 42-32=10

6 0

4 years ago

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Which is the additive inverse of the complex number -8+3i?

lawyer [7]

The additive inverse of a complex z is a complex number $-z,$ so that

$\mathsf{z+(-z)=-z+z=0+0i=0}$

Finding $-z:$

$\mathsf{z=-8+3i}\\\\\\ \mathsf{z+(-z)=0+0i}\\\\\ \mathsf{-8+3i+(-z)=0+0i}\\\\ \mathsf{-z=0+0i+8-3i}\\\\ \mathsf{-z=(0+8)+(0i-3i)}\\\\ \boxed{\begin{array}{c}\mathsf{-z=8-3i} \end{array}}\qquad\checkmark$

If you're having problems understanding this answer, try seeing it through your browser: <span>brainly.com/question/2159647

</span>I hope this helps.

Tags: <em>complex number additive inverse opposite algebra</em>

3 0

3 years ago

There are 27 pupils. There are twice as many boys than girls. how many boys and girls are there

Anna71 [15]

Let x = the numbers of girls
thus
the numbers of boys = 2x
x + 2x = 27
3x = 27
x = 9
2x = 2 . 9 =18
girls = 9 & boys = 18

3 0

3 years ago

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What’s equivalent to x^-2

lesantik [10]

1/x^2. Remember that any number raised to a negative power is the reciprocal.

5 0

4 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .

The area (in square inches) of the circle that the hexagon did not cover would be:

$\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}$ .

3 0

3 years ago