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Arada [10]
3 years ago
9

The area of a triangular block is 16 squre inches. If the base of the triangle is twice the height, how long are the base and th

e height of the triangle?
Mathematics
1 answer:
klemol [59]3 years ago
6 0
The area of a triangle can be calculated using the formula 

1/2 (base) (height)

Let's substitute our known values, letting x represent the value of the height of the triangle.

16  = 1/2* x* 2x

Let's Simplify!

16 = x^2

Finally, lets take the square root of both sides, to get rid of the exponent on the right side of the equation.

x= positive OR negative 4.

However, because x is equal to the height, we know that it can't be negative, so we know that it is positive 4.

The height of the triangle = 4 inches
The base of the triangle = 2h = 8 inches

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Melanie connected a brown garden hose, a green garden hose, and a black garden hose to make one long hose. The brown hose is 10.
natta225 [31]

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35.65 feet

Step-by-step explanation:

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3 0
3 years ago
Which is the graph of f(x) = -(x + 3)(x + 1)?
Bas_tet [7]

Answer:

See attachment

Step-by-step explanation:

A function is given to us and we need to tell which graph represents the given function. The function given to us is,

\tt: \implies f(x) = -( x + 3 )( x + 1 )

Let's find out at which points do the graph Intersects x axis / finding the roots. For that substitute f(x) = 0 , we have ,

\tt: \implies  -(x +3)( x + 1 ) = 0

Equate each factor by 0 ,

\tt: \implies \boxed{\blue{ \tt x = -1,-3 }}

Therefore the graph will intersect x axis at x is equal to -1 and x is equal to -3 .

On looking at the given graphs in the options the second graph intersects x axis at -1 and -3 .

<u>Hence</u><u> </u><u>the </u><u>second</u><u> </u><u>option</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u><u> </u>

{ See attachment }

4 0
3 years ago
Calculate this reflection of the triangle:
Mice21 [21]

Answer:

a = 3, b = 0, c = 0, d = -2

Step-by-step explanation:

<em>To find the reflection Multiply the matrices</em>

∵ The dimension of the first matrix is 2 × 2

∵ The dimension of the second matrix is 2 × 3

<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>

2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix

\left[\begin{array}{ccc}1&0\\0&-1\end{array}\right]  × \left[\begin{array}{ccc}0&3&0\\0&0&2\end{array}\right]  = \left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]

Compare the elements in the answer with the third matrix to find the values of a, b, c, and d

∴ a = 3

∴ b = 0

∴ c = 0

∴ d = -2

7 0
3 years ago
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