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Alex17521 [72]
2 years ago
6

Need Help....

Mathematics
2 answers:
natali 33 [55]2 years ago
5 0

Answer:

the \: expession \: is \: equivalent \: to \: 2

to \: know \: the \: solution \: with \: explanation

refer \: to \: the \: above \: attatchment

<h3>Carry on learning !! </h3>

Alecsey [184]2 years ago
3 0

If 0 < x < π/2, then both sin(x) and cos(x) are positive.

From the Pythagorean identity, we then have

cos²(x) + sin²(x) = 1

⇒   cos(x) = + √(1 - sin²(x))   and   sin(x) = + √(1 - cos²(x))

Then

√(1 - cos²(x))/sin(x) = sin(x)/sin(x) = 1

and

√(1 - sin²(x))/cos(x) = cos(x)/cos(x) = 1

so that the overall expression reduces to

1 - 1 = 0

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Determine the remainder when (x3 – 5x2 14x – 5) is divided by (x 3)
Bogdan [553]
Hello,

P(x)=x^3-5x^2+14x-5

P(-3)=(-3)^3-5*(-3)²+14*(-3)-5=-119
Remainder=-119
6 0
3 years ago
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If the radius of a circle is 15, what will be the area of the circle?
yarga [219]

Answer:

751.124

Step-by-step explanation:

area of a cricle is

πr^2

take π to be 22÷7

A= πr^2

A=22÷7*15*15

A=22÷7*225

7 divides 225

= 34.142

22*34.142

A= 751.124

8 0
2 years ago
As a single logarithm, 2log2 6- log29+ 1/3log2 27 = log2 12.
Ipatiy [6.2K]

Answer:

true

Step-by-step explanation:

2log2 6- log2 9+ 1/3log2 27 =

= log2 6² - log2 9 + log2 27^(1/3)

= log2 36 - log2 9 + log2 3

= log2 [(36 × 3)/9]

= log2 12

The statement is true.

3 0
2 years ago
12.28 as a mixed number
Bond [772]
12 28/100 , is the answer.
6 0
2 years ago
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If 180° &lt; α &lt; 270°, cos⁡ α = −817, 270° &lt; β &lt; 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?
eduard

Answer:

cos(\alpha+\beta)=-\frac{84}{85}

Step-by-step explanation:

we know that

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

Remember the identity

cos^{2} (x)+sin^2(x)=1

step 1

Find the value of sin(\alpha)

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

cos(\alpha)=-\frac{8}{17}

Find the value of sine

cos^{2} (\alpha)+sin^2(\alpha)=1

substitute

(-\frac{8}{17})^{2}+sin^2(\alpha)=1

sin^2(\alpha)=1-\frac{64}{289}

sin^2(\alpha)=\frac{225}{289}

sin(\alpha)=-\frac{15}{17}

step 2

Find the value of cos(\beta)

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

sin(\beta)=-\frac{4}{5}

Find the value of cosine

cos^{2} (\beta)+sin^2(\beta)=1

substitute

(-\frac{4}{5})^{2}+cos^2(\beta)=1

cos^2(\beta)=1-\frac{16}{25}

cos^2(\beta)=\frac{9}{25}

cos(\beta)=\frac{3}{5}

step 3

Find cos⁡ (α + β)

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

we have

cos(\alpha)=-\frac{8}{17}

sin(\alpha)=-\frac{15}{17}

sin(\beta)=-\frac{4}{5}

cos(\beta)=\frac{3}{5}

substitute

cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})

cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}

cos(\alpha+\beta)=-\frac{84}{85}

4 0
3 years ago
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