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Ira Lisetskai [31]

3 years ago

13

Help I’ve been working on this problem for about 2 days and I still don’t get it and please explain and show me how to do the wo

rk

1 answer:

Bond [772]3 years ago

4 0

the boiling point of jet fuel is 329 degrees F. Rounded to the nearest degree, what is the temperature in degrees Celcius? Use the Formula: C = 5/9(F - 32), where C represents degrees celcius, and F represents degrees fahrenheit.

A). 165 degrees C

B). 183 degrees C

C). 201 degrees C

D). 535 degrees C

Solve:

C = 5/9(F - 32)

Simplify both sides:

C = 5/9(329 - 32) = 0

Simplify:

5/9

Equation at end of step 1:

C - (5/9 * 297) = 0

C - 165 = 0

Solve single variable:

C - 165 = 0

Add 165 to both sides:

C = 165

Therefore, Your answer would be, Letter Choice (A), (C = 165 degrees C

Hope that helps!!!!! Have a great day!!!!! : )

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5. Combien de paires de nombres naturels peut-on former dont : a) la somme est 23? b) la différence est 100?

Lilit [14]

Answer: a) 24 b) infini

Step-by-step explanation: a) parce que tu peut faire que 24 somme pour faire 23. B) infini car tout plus haut 100 fonctione

3 0

1 year ago

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docker41 [41]

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Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

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

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

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

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

Name an example of an acute angle, right angle, and straight angle according to their angle measures

oksian1 [2.3K]

Acute: anything less than 90 but greater than 0

so, m<CAD

right: angles at exactly 90 degrees, so m<AEC,

and obtuse angles are anything greater than 90 but less than 180 degrees,

so CDA (you can tell by the picture that it is bigger than 90- 90 degrees is exactly perpendicular)

8 0

2 years ago

It is given that p varies inversely as q. If p = 12 and q = 45 find p, if q is 135.

alexgriva [62]

Answer:

$p =4$

Step-by-step explanation:

Given

Variation: Inversely

$p = 12$ when $q = 45$

Required

Determine p when $q = 135$

First, we need to determine the relationship between p and q

Since, it is an inverse variation.

The relationship is:

$p\ \alpha\ \frac{1}{q}$

Convert to an equation

$p= k *\frac{1}{q}$

$p= \frac{k}{q}$

Where k = constant of variation

Make k the subject

$k = p * q$

When $p = 12$ and $q = 45$

$k = 12 * 45$

$k = 540$

To solve for p when q = 135

Substitute 135 for q and 540 for k in $p= \frac{k}{q}$

$p = \frac{540}{135}$

$p =4$

6 0

2 years ago