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1 year ago

11

Find the volume of the triangular prism

2 answers:

juin [17]1 year ago

7 0

Answer:

Remember the formula for calculating volume is: Volume = Area by height. V = A X h.

For a triangle the area is calculated using the formula: Area = half of base by altitude. A = 0.5 X b X a.

So to calculate the volume of a triangular prism, the formula is: V = 0.5 X b X a X h.

Step-by-step explanation:

Alborosie1 year ago

7 0

Answer:

70

Step-by-step explanation:

5*4*7=140/2=70

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Llana [10]

Answer:

3x was subtracted from the left side, but 3x was subtracted from the right side. The Subtract Property of Equality states that you can subtract the same number from each side and the equation will remain true. But 3x and 3 are not the same number (unless x is 1).

x = -8/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

$(7x+3x)+(21)=5$
$10x + 21 = 5$

Step 2: Subtract 12 from both sides.

$10x + 21 - 21 = 5 - 21$
$10x = -16$

Step 3: Divide both sides by 10.

$\frac{10x}{10} = \frac{-16}{10}$
$x = -\frac{8}{5}$

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1) 2 > -3 > -5

Step-by-step explanation:

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2 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

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2 years ago

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