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3 years ago

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Please help this is urgent and it's due today.

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Llana [10]3 years ago

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What is ?
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Show your work with long division please

Naya [18.7K]

Answer:

<h3> __</h3><h3>0.63</h3>

Step-by-step explanation:

7/11 = no, add 0

70/11 = 6, so 0.6, remainder is 4, add 0

40/11 = 3, so 0.03, remainder is 7, add 0

70/11 = 6... and it goes on

<h3>The answer is 0.63 bar notation on both 6 and 3</h3>

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3 years ago

To date, this year, Company XYZ has sold 450 units. They sell units at an average rate of 15 per week. The company wants to sell

Arisa [49]

The inequality used to solve for x is x=750/15

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3 years ago

I suck at math don't judge.. 9/8 divided by 3/6 =? i got an answer just don't know if i'm correct

olga nikolaevna [1]

Its 2.25 I believe ..............

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3 years ago

A pair of vertical angles have measures (2z+43)° and (−10z+25)° .<br><br> What is the value of z?

Alexxx [7]

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3 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

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