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2 years ago

What is the area of this circle? The circle is 70 m

Mathematics

1 answer:

Anni [7]2 years ago

3 0

The answer is 34<span> because all you do is go from the radius to the diameter.

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Nicholas wrote the steps below to simplify the fraction 20/30. Find his error and correct it. 20/30= 20÷5=4 30÷6=5

aksik [14]

Answer:2/3

Step-by-step explanation:

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2 years ago

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Help I’m really bad at this

kogti [31]

Answer:

Step-by-step explanation:

The formula for surface area is SA = 2lw + 2wh + 2lh

W = width

L= length

H = height

A = 2(wl + hl + hw)

2·(6·3+2·3+2·6)

Simplify that down to get the answer 72

4 0

3 years ago

M is directly proportional to r2 <br> When r=2, m=14<br> Work out the value of m when r=12

Genrish500 [490]

we have M is durectly porpotional to r^2

so M=(k)r^2

and when r=2, m=14

so 14=(k)(2)^2

k=14/4 =7/2

so when r=12

m= (7/2)(12)^2 =(7/2)(144) = 504

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2 years ago

Select the two values of x that are roots of this equation.

alina1380 [7]

C
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2 years ago

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30

Dafna11 [192]

Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

$\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}$

$\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------$

$\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o$

8 0

2 years ago