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2 years ago

7

(10pts) find all the solutions to the equations.

2 answers:

kari74 [83]2 years ago

8 0

I think they are all correct

xxTIMURxx [149]2 years ago

4 0

All of your solution is right

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A triangle has a 60° angle, and the two adjacent sides are 12 and "12 times the square root of 3". Find the radius of a circle w

shusha [124]

The radius of a circle with the same vertex as a center is 12 units

<h3>Application of Pythagoras theorem;</h3>

To get the radius of the circle, we need to determine the diameter of the circle first:

According to SOH CAH TOA:

$sin\theta = \frac{opp}{hyp} \\sin60 = \frac{12\sqrt{3}}{hyp} \\hyp =\frac{12\sqrt{3}}{sin60} \\hyp =\frac{2\times12\sqrt{3}}{\sqrt3}} \\hyp = 24 = diameter$

Determine the radius of the circle

Radius = dismeter/2

Radius = 24/2

Radius = 12

Hence the radius of a circle with the same vertex as a center is 12 units

Learn more on radius of a circle here: brainly.com/question/24375372

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2 years ago

What is the volume of the pyramid

alexgriva [62]

Answer:

Option A - 32 cm³

Step-by-step explanation:

Volume of a pyramid is given by the formula :

V = $\frac{lwh}{3}$

Substituting values given gives us :

V = (4)(4)(6)/3

V = (16)(6)/3

V = 96/3

V = 32

Units will be cm³ as this is volume

Hope this helped and have a good day

5 0

1 year ago

Read 2 more answers

Only question numer 10.

leva [86]

Out of the six, only “u” wouldn’t work so the answer is d. 5

Hope this helps and hope you have a great day! Please make brainiest

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2 years ago

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X^(log_(\sqrt(x))(x-3))=4

Zina [86]

Answer:

X= -4

Step-by-step explanation:

3 0

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

8 0

3 years ago

Read 2 more answers