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leonid [27]
3 years ago
11

A solid right pyramid has a square base. The length of the

Mathematics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

The volume of the pyramid is 16 cm³.

Step-by-step explanation:

The volume of squared-base pyramid is given by the formula:

V=A\times \frac{h}{3}

Here,

V = volume of the squared-base pyramid

A = area of the square base

<em>h</em> = height of the pyramid.

The information provided is:

<em>a</em> = side of square = 4 cm

<em>h</em> = 3 cm

Compute the area of the square base as follows:

A=a^{2}\\=4^{2}\\=16\ \text{cm}^{2}

Compute the volume of squared-base pyramid as follows:

V=A\times \frac{h}{3}

   =16\times \frac{3}{3}\\\\=16

Thus, the volume of the pyramid is 16 cm³.

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Find the product using the number line. 3(-6) = A number line going from negative 7 to positive 3. An arrow goes from negative 6
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Step-by-step explanation:

B - The arrows should be length of 6.

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E - The arrows should point in the negative direction

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Please verify the following trigonometric identities.
Nina [5.8K]

Seems like there is a correction in the first question (RHS is tanx.tany)

(i) For convenience: let tanx = a ; tany = b

Thus, tanx + tany = a + b

Moreover, cotx = 1/tanx = 1/a ; coty = 1/b

Thus,

cotx + coty = 1/a + 1/b = (a + b)/ab

Hence,

=> (tanx + tany)/(cotx + coty)

=> (a + b) / { (a + b)/ab }

=> ab(a + b)/(a + b)

=> ab => tanx.tany , proved.

(ii) For convenience: let sinx = a ; cosx = b

As we know, sin²x + cos²x = 1 => a²+b²=1

=> (a³ + b³)/(a + b)

=> (a + b)(a² + b² - ab) / (a + b)

=> (a² + b² - ab)

=> 1 - ab => 1 - sinx.cosx , proved

(iii): let x/2 = A

=> tan(x/2) + cosx.tan(x/2)

=> tanA + cos2A.tanA

=> tanA [1 + cos2A]

=> tanA (2cos²A) {1+cos2A = 2cos²A}

=> (sinA/cosA) (2cos²A)

=> sinA (2cosA)

=> 2sinAcosA

=> sin2A

=> sin2(x/2)

=> sinx proved

Letting x/2 = A is not mandatory. I did it to decease words*(in a line).

<u>Indentities used</u>:

• sin²A + cos²A = 1

• (a³ + b³) = (a + b)(a² + b² - 1)

• 1 + cosA = 2 cos²(A/2)

• tanA = sinA/cosA.

• 2sinAcosA = sin2A

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What is a solution to the system of equations that includes quadratic function f(x) and linear function g(x)? f(x) = 2x2 + x + 4
olga_2 [115]

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is

(\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

We have given that,

f(x) = 2x^2 + x + 4

x                          g(x)      

-2                               1      

  -1                              3        

  0                              5      

  1                              7

  2                              9

<h3>What is a polynomial function?</h3>

A polynomial function is a relation where a dependent variable is equal to a  polynomial expression.

A polynomial expression is an expression including numbers and variables, where variables are raised to non-negative powers.

The general form of a polynomial expression is:

a₀ + a₁x + a₂x² + a₃x³ + ... + anxⁿ.

The highest power to a variable is the degree of the polynomial expression. When degree = 2, the function is a quadratic function.

When degree = 1, the function is a linear function.

<h3>How do we solve the given question?</h3>

The quadratic function is given to us:f(x) = 2x^2 + x + 4.

We need to determine the linear equation g(x).

Since it's a linear equation we use the two-point method to determine the equation.

<h3>What is the two-point?</h3>

y-y₁ = ((y₂-y₁)/(x₂-x₁))*(x-x₁)

We take the points g(-2) =1, g(-1) = 3

g(x) - g(1) = ((g(-2)-g(-1))/(-2+1))*(x-1)or,

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g(x) = x - 1 + 1 = x

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For that, we equate f(x) and g(x).2x^2 + x + 4 = x or, 2x² - x +x- 4 = 0or, 2x^2  - 4 = 02(x^2-2)=0x^2-2=0x^2-\sqrt{2}=0(x-\sqrt{2}) (x+\sqrt{2})=0(x-\sqrt{2})=0 \\x=\sqrt{2}  or x=-\sqrt{2}g(-1) = 3(from the table)g(\sqrt{2})=\sqrt{2}  \ and \ g(\sqrt{-2}) =-\sqrt{2}f(\sqrt{2}) = 2(\sqrt{2} )^2 + (\sqrt{2} ) + 3\\=2(2)+\sqrt{2} +3\\=7+\sqrt{2}g(-\sqrt{2} )= 2(\sqrt{-2} )^2 + (\sqrt{-2} ) + 3\\=2(2)+\sqrt{-2} +3\\=7+\sqrt{-2}

The solution to the system of equations that includes quadratic function f(x) and linear function g(x) is (\sqrt{2},\sqrt{-2}  ) and ((7+\sqrt{2)} ,(7+\sqrt{-2}))

Learn more about linear and  quadratic equations at

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