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harina [27]
3 years ago
6

A rectangle has an area of 24cm, how long could the sides of the rectangle be? Give 3 examples,

Mathematics
1 answer:
Flura [38]3 years ago
3 0
1. 12cm long and 2cm wide

2. 8cm long and 3cm wide

3. 6cm long and 4cm wide

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Greatest 22,17,1,-10,-11,-26 least
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Which of these shows the following expression factored completely? 6x2 – 13x +5 ​
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B

Step-by-step explanation:

That's my guess.

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What is 78.6 in word form
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78.6 = Seventy eight and six tenths.
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Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
arsen [322]
Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

3 0
3 years ago
Write a quadratic function with zeroes 0 and 8.
Jlenok [28]

Answer:

x^2-8x=0

Step-by-step explanation:

the quadratic function should be as follows:

x^2-8x=0

Now let's confirm that the zeros of the function are 0 and 8

x^2-8x=0=x(x-8)=0

Therefore we can see that if x = 0

0^2*8*0=0\\0=0

the equation is fulfilled

And we also have (x-8)

for this expresion to be equal to zero:

x-8=0\\x=8

thus, if x = 8

8^2-8*8=0\\64-64=0\\0=0

the equation is also fulfilled

The zeros of the quadratic function x^2-8x=0 are 0 and 8.

3 0
3 years ago
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