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3 years ago

What is the simplest form for 13/8

Mathematics

2 answers:

Crank3 years ago

6 0

Its already in simplest form but you can convert it to a mixed fraction and get:
1 5/8

velikii [3]3 years ago

3 0

13/8=1 5/8= 1.625

Hope this helps!!!

Brainliest Please

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from a vertical cliff 80 m above sea level a fishing boat is observed at an angle of depression of 6 degrees . how far out to se

san4es73 [151]

Answer:

The boat is 8.4m out of the sea

Step-by-step explanation:

Kindly find attached rough sketch of the problem.

From the sketch we can see that the observer from the cliff and the horizontal distance forms a right angle triangle

Let the horizontal distance be x

Let the angle of elevation from the boat be y

Applying sum of angles in a triangle

90+6+y= 180

96+y=180

y=180-96

y= 84°

Applying SOH CAH TOA

Using TOA

Tanθ= opposite /adjacent

Tan 84= 80/x

x= 80/tan 84

x= 80/9.51

x= 8.4m

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4 years ago

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4 years ago

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den301095 [7]

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3 years ago

If the sinø=7/12, what is cosø?

777dan777 [17]

Answer:

If the sinø=7/12, cosø is $\frac{\sqrt{95} }{12}$

Step-by-step explanation:

We are given sinФ = 7/12

We need to find cosФ

The basic trigonometric functions of right triangle are:

$sin\theta=\frac{opposite}{hypotenuse}$

$cos\theta=\frac{adjacent}{hypotenuse}$

We need values of adjacent and hypotenuse to find cosФ

Using Pythagoras theorem we can find the value of adjacent

$(Hypotenuse)^2= (Opposite)^2+(Adjacent)^2$

We have Hypotenuse= 12 and Opposite = 7 (because $sin\theta=\frac{opposite}{hypotenuse}$ and we are given $sin\theta=\frac{7}{12}$

Inserting values and finding adjacent:

$(Hypotenuse)^2= (Opposite)^2+(Adjacent)^2\\(12)^2=(7)^2+(Adjacent)^2\\144=49+(Adjacent)^2\\144-49=(Adjacent)^2\\95=(Adjacent)^2\\\sqrt{(Adjacent)^2} =\sqrt{95}\\Adjacent=\sqrt{95}$

So, value of Adjacent is $\sqrt{95}$

Now finding cosФ

$cos\theta=\frac{adjacent}{hypotenuse}$

Adjacent = $\sqrt{95}$ , hypotenuse = 12

$cos\theta=\frac{\sqrt{95} }{12}$

So, If the sinø=7/12, cosø is $\frac{\sqrt{95} }{12}$

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sukhopar [10]

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