Which expression is equivalent to squareroot 36/25

Lera25 [3.4K]

Answer:

-11 + 4

Step-by-step explanation:

When you subtract a negative the sign will switch to plus, so it this case you would switch the - (-4) to a + 4 but you don't change the -11 at the front.

6 0

4 years ago

Suppose we have a right triangle with legs of length a and b and hypotenuse of length c. Suppose b=3 and c=5. Then a= , For the

ANTONII [103]

Answer:

Length of right-angle triangle 'a' = 4

b)

$sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$

$cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$

$tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$

Step-by-step explanation:

Step(i):-

Given b = 3 and hypotenuse c = 5

Given ΔABC is a right angle triangle

By using pythagoras theorem

c² = a² + b²

⇒ a² = c² - b²

⇒ a² = 5²-3²

=25 - 9

a² = 16

⇒ a = √16 = 4

The sides of right angle triangle a = 4 ,b = 3 and c = 5

Step(ii):-

$sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$

$cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$

$tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$

7 0

3 years ago

Line A passes through the points (1, -1) and (-2, 8).

Viefleur [7K]

The answer to this question will be D

5 0

3 years ago

Pleaseeee i don’t understand!

zhannawk [14.2K]

Answer:

its april fools

Step-by-step explanation:

5 0

3 years ago

Solve sin4xcos2x-cos4xsin2x=square root of 2 sinx over the interval [0,2pi)

ddd [48]

$\bf sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})
\\\\\\
sin(2\theta)=2sin(\theta)cos(\theta)\\\\
-------------------------------\\\\
sin(4x)cos(2x)-cos(4x)sin(2x)=\sqrt{2}sin(x)
\\\\\\
sin(4x~~-~~2x)=sin(x)\sqrt{2}\implies sin(2x)=sin(x)\sqrt{2}
\\\\\\
$

$\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0
\\\\\\
sin(x)~[2cos(x)-\sqrt{2}]=0\\\\
-------------------------------\\\\
sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\
-------------------------------\\\\
2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2}
\\\\\\
\measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}$

now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).

7 0

4 years ago