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3 years ago

A playground has the shape of a trapezoid.

Mathematics

1 answer:

Dmitrij [34]3 years ago

6 0

Answer:

short base: 6 yards
long base: 8 yards

Step-by-step explanation:

Our understanding of your figure is shown below.

The question says the "shortest side" and the "width" have the same dimension. If the "width" is a reference to "height 6 yards", then it seems the "shortest side" is 6 yards. Since the slant sides are longer than the height, the "shortest side" is also the "short base."

The short base is 6 yards.

The long base overhangs the short base by 1 yard on either end, so it is a total of 2 yards longer than the short base. It it 6+2 = 8 yards long.

The long base is 8 yards.

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How many variable terms are in the expression 3x3y + 5xz − 4y + x − z + 9?

Dimas [21]

Answer:

Step-by-step explanation:

there are only x y and z used in the expression no matter how many times they multiply.

8 0

2 years ago

Help me out asap!!!!

attashe74 [19]

Answer:

anglePUM = 3° , PM = 5, UM = √75

Step-by-step explanation:

anglePUM = 180 - 60 - 90 = 30° (∠sum of Δ)

cos60 = PM/10

PM = 10cos60 = 5

sin60 = UM/10

UM = 10sin60 = √75

6 0

2 years ago

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

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