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irina1246 [14]
3 years ago
7

Factor 49a2 + 7a - 6.

Mathematics
1 answer:
Oxana [17]3 years ago
7 0
The answer of the question is B
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Vance is designing a garden in the shape of an isosceles triangle. The base of the garden is 30 feet long. The function y = 15 t
Sever21 [200]

Answer:

Part 1) The height of the triangle when θ = 30° is equal to 8.66\ ft

Part 2) The height of the triangle when θ = 40° is equal to 12.59\ ft

Part 3) The area of triangle with θ = 30° is less than the area of triangle with θ = 40°

Step-by-step explanation:

Part 1) What is the height of the triangle when θ = 30 ° ?

we have

y=15tan(\theta)

substitute the value of theta in the equation and find the height

y=15tan(30\°)=8.66\ ft

Part 2) What is the height of the triangle when θ = 40 ° ?

we have

y=15tan(\theta)

substitute the value of theta in the equation and find the height

y=15tan(40\°)=12.59\ ft

Part 2) Vance is considering using either θ = 30 ° or θ = 40 ° for his garden

Compare the areas of the two possible gardens

step 1

Find the area when θ = 30 °

The height is 8.66\ ft

Remember that the area of a triangle  is equal to the base multiplied by the height and divided by two

so

A=(1/2)(30)(8.66)=129.9\ ft^{2}

step 2

Find the area when θ = 40°

The height is 12.59\ ft

Remember that the area of a triangle  is equal to the base multiplied by the height and divided by two

so

A=(1/2)(30)(12.59)=188.85\ ft^{2}

Compare the areas of the two possible gardens

The area of triangle with θ = 30° is less than the area of triangle with θ = 40°

7 0
3 years ago
Can anyone help me do this please? No Links!!
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$100,500............

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Step-by-step explanation:

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3x+6/x^2-x-6+2x/x^2+x-12
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\dfrac{3x+6}{x^2-x-6} +  \dfrac{2x}{x^2+x-12}

= \dfrac{3(x+2)}{(x + 2)(x - 3)} +  \dfrac{2x}{ (x - 3)(x + 4)}

= \dfrac{3}{(x - 3)} +  \dfrac{2x}{ (x -3)(x + 4)}

= \dfrac{3(x+4)}{(x - 3)(x + 4)} +  \dfrac{2x}{ (x -3 )(x + 4)}

= \dfrac{3(x+4) + 2x }{(x - 3)(x + 4)}

= \dfrac{3x+12 + 2x }{(x - 3)(x + 4)}

= \dfrac{5x+12}{(x - 3)(x + 4)}



6 0
3 years ago
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