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MariettaO [177]

2 years ago

13

Can someone help me with this problem!

1 answer:

Juli2301 [7.4K]2 years ago

3 0

-5 I think

Ignore this this is just for 20 words

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Select all that are true

sertanlavr [38]

A=-1, b=2 I hope this helped. Please please please if you like my answer mark Brainly:)

3 0

1 year ago

At combs store there is a 25% chance a customer enters the store within one minute of closing time. Describe the complementary e

12345 [234]

The complementary event is that a customer does not enter the store within one minute of closing, and the probability is 75%.

The complement is found by subtracting from 100%; 100-25 = 75%.

6 0

3 years ago

Solve to indicated variable 16ak=-5 for a

damaskus [11]

The answer is
$\frac{ - 5}{16k} = a$
steps
#1) Try to leave a by itself by dividing any variable next to it
$\frac{16 ak = - 5 }{ 16k}$
you can only divided these variables as all these variables in the original equation. are being multiplied. The opposite if multiplication is division.

3 0

3 years ago

If 4/3divided by 1/6 then the value of p is between which of the following pairs of numbers

Lerok [7]

Answer:

7 and 9

Step-by-step explanation:

4/3÷1/6 = 4/3 × 6/1

= 8

So the value of p falls between the range of 7 and 9

3 0

2 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

2 years ago