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Vedmedyk [2.9K]

2 years ago

10

The numerator of the expression 3-x/x-3 can be written as

1 answer:

Ierofanga [76]2 years ago

3 0

Answer:

The answer is -(x-3)

Step-by-step explanation:

It can be written as a negative number so it can be divided/cut :

3 - x = -x + 3

= -(x+3)

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Centroid of triangle with vertices (0,0) ,(3,0) & (0,-3) is……. [ ]

Firdavs [7]

Answer:

b

Step-by-step explanation:

centroid

x=(x1+x2+x3)/3=(0+3+0)/3=3/3=1

y=(y1+y2+y3)/3=(0+0-3)/3=-3/3=-1

centroid =(1,-1)

3 0

2 years ago

Read 2 more answers

How much greater is the area of a square with a side length of 9 inches than the area of a circle with a radius of 3 inches?

fredd [130]

The area of the square is 81 sq in, and the area of the circule is (3.14)(3 in)^2, or approx 3.14(9) sq in, or approx 28.27 sq in.

Subtract: 81 sq in - 28.27 sq in = approx. 52.73 sq in.

The difference is 52.73 sq in; the square is larger, the circle smaller.

7 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

What free resources would you recommend for assistance in maths?

Andru [333]

Usually bbc bitesize is what I use for basically everything. Here’s a link:
https://www.bbc.com/bitesize

3 0

3 years ago

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Please solve numbers 16-17

Gelneren [198K]

Answer:

-1 :)

Step-by-step explanation:

4 0

2 years ago