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3 years ago

9

The number at the bottom of the exponent is called a power, true or false

2 answers:

Anna007 [38]3 years ago

7 0

Answer:

false

Step-by-step explanation:

the number at the bottom of the exponent is called a base. for example, if your exponent is 2^3, as shown in the image, the 2 would be the base and the 3 would be the power :)

SOVA2 [1]3 years ago

4 0

Answer:

False.

Step-by-step explanation:

Let's say we had $2^{4}$ we would say 2 to the power of 4.

2 is the base and 4 is the exponent.

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The sides of a triangle are in the ratio 12:17: 25 and its perimeter is 540cm. The area is:

Feliz [49]

1000

Step-by-step explanation:

Find the height of the triangle using the Pythagoras theorem

Then, use it to calculate the area of the triangle

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3 years ago

Which is an example of an algebraic expression?

azamat

An algebraic expression doesnt have an equals sign or an inequality sign.

So,

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3 years ago

How many presents are between 40_45 years old

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2 years ago

Read 2 more answers

Give me an example of an irrational number that is greater than 10

Pavel [41]

To find:

An irrational number that is greater than 10.

Solution:

Irritation number: It cannot be expression in the form of $\dfrac{p}{q}$ , where, $q\neq 0, p,q$ are integers.

For example: $\sqrt{2}\sqrt{3},\pi, 1.263689...,etc.$ .

We know that square of 10 is 100. So, square root of any prime number is an example of an irrational number that is greater than 10.

First prime number after 100 is 101.

Required irrational number $=\sqrt{101}$

Therefore, $\sqrt{101}$ is an irrational number that is greater than 10.

5 0

3 years ago

Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago