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3 years ago

6

Which is greater 1/4 or 0.4

2 answers:

ahrayia [7]3 years ago

6 0

Answer:

0.4 is greater than 1./4.

Likurg_2 [28]3 years ago

3 0

Answer:

0.4

Step-by-step explanation:

1/4 = 0.25

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A parallelogram has sides 17.3 m and 43.4 m long. The height corresponding to the 17.3-m base is 8.7 m. Find the height, to the

Anvisha [2.4K]

Answer:

3.5m

Step-by-step explanation:

A parallelogram has sides 17.3 m and 43.4 m long.

The height corresponding to the 17.3-m base is 8.7 m.

The area of a parallelogram = base × height

= 8.7m × 17.3m

= 150.51m²

Both parallelograms have the same area

Hence, the height, to the nearest tenth of a meter, corresponding to the 43.4-m base is calculated as:

= 150.51m²/43.4m

= 3.4679723502m

Approximately = 3.5m

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2 years ago

Here is 100 brain coins have a good day

Harrizon [31]

Answer:

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2 years ago

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

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

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Lisa says that 43 is an odd number that is composite . Is she correct? Explain.

Assoli18 [71]

Lisa is incorrect although it is an odd number it is not composite. It only has itself and 1. Therefore its a prime number.

This is the definition for prime if you still dont understand.

A prime number is a whole number greater than 1 whose only factors are 1 and itself.

6 0

3 years ago

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Robin has a table full of supplies that she will use to decorate picture frames. She decorated one frame with 3 buttons and 12 b

TiliK225 [7]

The ratio is 1:4
She could use 4 gems and 16 flowers 4 x 4 = 16 7 stickers and 28 glue dots 7 x 4 = 28

8 0

3 years ago