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kodGreya [7K]
4 years ago
10

Tell whether the triangle with the given side lengths is a right triangle.

Mathematics
1 answer:
harina [27]4 years ago
3 0

Answer:

It is a right triangle

Step-by-step explanation:

Let,

Base of the triangle = b = 10.5 feet

Hypotenuse of the triangle= h = 14.5 feet

Perpendicular of the triangle = p = 10 feet

According to Pythagoras  theorem;

h^{2}=p^{2} +b^{2}

14.5^{2}=10^{2} +10.5^{2}

210.25=100+110.25

210.25=210.25

As both side is equal, so it is a right triangle.

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I’m confused on c and d please help me
Rudik [331]

Answer:

C is 27 and d is 16

Step-by-step explanation:

count how many cubes are on the first level and multiple but how many it goes up too

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4 0
3 years ago
3.
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
3 years ago
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