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3 years ago

Please find the perimeter!

Mathematics

1 answer:

puteri [66]3 years ago

7 0

Answer:

1. half circle: C = (2πr) / 2

a. C = (2π4) / 2 ≈ 12.566

2. triangle:

a. find slant / hypotenuse.

i. 4² + 10² = c²

ii. 16 + 100 = c²

iii. c² = 116

iv. c ≈ 10.770

b. add both slants to find triangular area

i. 10.77 + 10.77 ≈ 21.54

3. add:

a. 12.566 + 21.54 ≈ 34.106 cm

hope this helps :)

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natima [27]

Answer:

See explanation

Step-by-step explanation:

Let x be the number of spade shovels, y -the number of flat shovels and z - the number of square showels sold that day.

The store keeps an inventory of 80 shovels, then

x+y+z=80

The store always buy twice as many spade shovels as square, so

x=2z

The total cost of all shovels is

16x+9.60y+12.80z=1,072

a) The system of three equations is

$\left\{\begin{array}{l}x+y+z=80\\ \\x=2z\\ \\16x+9.60y+12.80z=1,072\end{array}\right.$

b) In matrix form this is

$\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}80\\0\\1,072\end{array}\right)$

c) The determinant is

$\left\|\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right\|=0-32+9.60-0+19.20-12.80=-16$

d) Find three determinants:

$\left\|\begin{array}{ccc}80&1&1\\ 0&0&-2\\ 1,072&9.60&12.80\end{array}\right\|=0-2,144+0-0+1,536-0=-608$

$\left\|\begin{array}{ccc}1&80&1\\ 1&0&-2\\ 16&1,072&12.80\end{array}\right\|=0-2,560+1,072-0+2,144-1,024=-368$

$\left\|\begin{array}{ccc}1&1&80\\ 1&0&0\\ 16&9.60&1,072\end{array}\right\|=0+0+768-0-0-1,072=-304$

So,

$x=\dfrac{-608}{-16}=38\\ \\y=\dfrac{-368}{-16}=23\\ \\z=\dfrac{-304}{-16}=19$

e) If the store doubled all prices and inventory, then the new matrix is

$\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 32&19.20&25.60\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}160\\0\\2,144\end{array}\right)$

7 0

2 years ago

I need to find the limit

Digiron [165]

For $x=\dfrac{1}{2}$ the function is undefined.

$\displaystyle
\lim_{x\to\tfrac{1}{2}^-} x\sec (\pi x)=\infty\\
\lim_{x\to\tfrac{1}{2}^+} x\sec (\pi x)=-\infty\\\\
\lim_{x\to\tfrac{1}{2}^-}x\sec (\pi x)\not = \lim_{x\to\tfrac{1}{2}^+}x\sec (\pi x)
$
So, the limit doesn't exist.

8 0

2 years ago

Please help WILL MARK BRAINLIEST HURRY LOOK AT PHOTO

Aleks04 [339]

Answer:

I believe it is D I'm a senior in pre-cal but I also dont have paper so I did it in my head rq

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2 years ago

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lord [1]

Answer: