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3 years ago

6

What statement explain whether haruto is correct

2 answers:

EastWind [94]3 years ago

6 0

Answer:

"Haruto is correct because the angle is coterminal with 3pi/4 and the reference angle is pi/4."

Step-by-step explanation:

Coterminal angle:

Coterminal Angles are angles who share the same initial side and terminal sides. Finding coterminal angles is as simple as adding or subtracting 360° or 2π to each angle, depending on whether the given angle is in degrees or radians.

Finding if 3pi/4 is coterminal with 19pi/4:

19pi/4

= 16pi/4 + 3pi/4

= 4pi + 3pi/4

= 2(2pi) + 3pi/4

= 2(360°) + 3pi/4

= 3pi/4

Hence they are coterminal.

Reference angle:

The smallest angle that the terminal side makes with x-axis.

3pi/4 falls in second quadrant.

In 2nd quadrant, reference angle is given as = pi - 3pi/4

reference angle = pi/4.

miv72 [106K]3 years ago

4 0

Answer:

"Haruto is correct because the angle is coterminal with 3pi/4 and the reference angle is pi/4."

Step-by-step explanation:

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LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

3 years ago