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4 years ago

The task is in the attachment, thank you

Mathematics

1 answer:

Novosadov [1.4K]4 years ago

4 0

Answer:

m=0

Step-by-step explanation:

mx²+2x-1=0

if x=1/2 then

m(1/2)² +2(1/2)-1=0

m/4+1-1=0

m/4=0

m=0

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago

Find the volume of the prism. 3 cm 37 cm 5 cm

vitfil [10]

The answer to your question is 45

5 0

2 years ago

What is the solution to x - 9 > -15? x > -6 x -24 x < -24

DaniilM [7]

Answer:

x > -6

Step-by-step explanation:

X - 9 > -15

x > -6

4 0

3 years ago

Read 2 more answers

Help needed ASAP will give you BRAINLIEST and 5 stars rate

Gwar [14]

Answer:

1 which is B

Step-by-step explanation:

Basically you just miltiply 2 * 0.5 which gives u 1. Hope this helps.

4 0

3 years ago

1. Josefina ha pintado 2/4 de una pared y su hermano  ¿Cuánto han pintado entre los dos?

Irina18 [472]

Answer 67+67=6886

Step-by-step explanation:

8 0

3 years ago

The task is in the attachment, thank you​

The task is in the attachment, thank you