Ask question

Ask question

All categories

4 years ago

13

Find the slope of the line that passes through (8, 4) and (5, 2).

1 answer:

Troyanec [42]4 years ago

8 0

If u you do the slope formula...(m=y2-y1/x2-x1)

2-4/5-8
-2/-3=2/3
Therefore the slope would be 2/3
Btw can u plz mark this answer the brainliest

You might be interested in

What is an Irrational Number?<br> Your answer<br> I

motikmotik

Answer:

An irrational number is a number that has no pattern and has no end. For example 3 divided by 7 is an irrational number because of the lack of pattern and never-ending decimal.

4 0

3 years ago

Read 2 more answers

What the square root of 49

luda_lava [24]

Answer:

7

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

Determine whether the statement is sometimes, always, or never true. If ax+b-4=b and doesn’t equal 0 then x=4/a?

attashe74 [19]

Start with expression $ax+b-4=b.$

1. Separate terms - with x leave in left side, without x write in right side:

$ax=b-b+4,\\ \\ax=4.$

2. Divide the last expression by a ( $a\neq 0$ ):

$x=\dfrac{4}{a}.$

Answer: this statement is true when $a\neq 0$ .

4 0

3 years ago

What error did katerina likely make?

Stella [2.4K]

Answer:

Katerina took the reciprocal of y/x

8 0

3 years ago

Read 2 more answers