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3 years ago

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A gemstone is cut in the shape of a triangular pyramid. The area of the base of the pyramid is 42 square millimeters. The height

of the pyramid is 4 millimeters. What is the volume of the gemstone? (Recall the formula V = one-third B h.)

1 answer:

never [62]3 years ago

7 0

Answer:

the volume of the gemstone is 56 millimeters^3

Step-by-step explanation:

The computation of the volume of the gemstone is shown below:

As we know that

The volume of the triangular pyramid is

= (1 ÷ 3) × (base area) × h

Now put the given values to the above formula

= 1 ÷ 3 × 42 × 4

= 56 millimeters^3

Hence, the volume of the gemstone is 56 millimeters^3

The above formula should be used and the same should be considered

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Solve using substitution x=(3y+1) and x+4y=15

natka813 [3]

The solution is: x= 7 and y = 2

Step-by-step explanation:

Given equations are:

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Putting x = 3y+1 in equation 2 from equation 1

$3y+1+4y = 15\\7y+1 = 15\\$

Subtracting 1 from both sides

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Keywords: Linear equations, substitution method

Learn more about linear equations at:

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3 years ago

I need to know how to find this.

GalinKa [24]

Answer:

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Step-by-step explanation:

When two parallel lines are crossed by another line (the transversal), the angles in matching corners are called corresponding angles. DCF and GFC isn't the correct answer because those would be vertical angles. GFC and BCA would not be corresponding because they do not fall in matching corners. And same with EFH and BCA. Therefore, BCF and EFH would be corresponding.

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3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

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3 years ago