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2 years ago

11

Whole Lot If there are 7 bears in a room and they each hug each other once and only once, how many total bear hugs were there al

l together?

1 answer:

Dvinal [7]2 years ago

3 0

1.They basically all did one big bug,not separate hugs with different bears.

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Theresa has two brothers,Paul and Steve, who are both the same height. Paul says he is 16 inches shorter than 1 1/2 times Theres

shepuryov [24]

Answer:

Theresa's height is 60 inches

Step-by-step explanation:

The given parameters are;

The height of Paul = The height of Steve

Paul's height = $1\dfrac{1}{2}$ times Theresa's height - 16 inches

Steve's height = $1\dfrac{1}{3}$ times Theresa's height - 6 inches

Let 'T' represents Theresa's height, we have;

Paul's height = $1\dfrac{1}{2}$ ×T - 16

Steve's height = $1\dfrac{1}{3}$ ×T - 6

Paul's height = Steve's height

Therefore;

$1\dfrac{1}{2}$ ×T - 16 = $1\dfrac{1}{3}$ ×T - 6

$1\dfrac{1}{2}$ ×T - $1\dfrac{1}{3}$ ×T = 10

$#\dfrac{3}{2}$ ×T - $\dfrac{4}{3}$ ×T = 10

$\dfrac{1}{6}$ ×T = 10

T = 10 × 6 = 60

Theresa's height, T = 60 inches

5 0

3 years ago

At prom, Dina asks a boy to dance. A moment later, her friend decides to ask a boy to dance, too.

leonid [27]

Answer:

Dina's friend is jealous of Dina's confidence so she wanted to "what up" her :)

Step-by-step explanation:

3 0

3 years ago

A semicircle has an area of 76.97 squared km. Find it’s radius

shtirl [24]

Answer:

7 km

Step-by-step explanation:

Substitute:

The equation for area of a semi-circle is $\frac{r^{2}\pi }{2}$

A = $\frac{r^{2}\pi }{2}$

76.97 = $\frac{r^{2}\pi }{2}$

Multiply by 2 on both sides:

2(76.97) = ( $\frac{r^{2}\pi }{2}$ )2

153.94 = $r^{2} \pi$

Divide by pi on both sides:

153.94 = $r^{2} \pi$

/pi /pi

49.03 ≈ $r^{2}$

Find the square root:

$\sqrt{49.03} =\sqrt{r^{2} }$

7 = r

7 0

3 years ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago

Which equation matches the function shown in the graph?

borishaifa [10]

Answer:

its c ! just took the test :)

5 0

3 years ago