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3 years ago

Please help me, if you help then thanks!

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

4 0

Answer:

1 / 2 is one side of the cube

Step-by-step explanation:

Take the cube root of 1/8 by taking the cube root of 1 which is 1 and then 8 which is 2 so equal to 1/2.

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14x = -7y + 21 Write in standard form.

Nuetrik [128]

Answer:

14x + 7y = 21 is your answer.

Step-by-step explanation:

Standard form is Ax + By = C

Isolate the constant. Note the equal sign, what you do to one side, you do to the other. Add 7y to both sides.

14x = -7y + 21

14x (+7y) = -7y (+7y) + 21

14x + 7y = 21

14x + 7y = 21 is your answer.

8 0

3 years ago

Read 2 more answers

Which trigonometric ratios are correct for triangle ABC? Check all that apply.

Alina [70]

Assuming the - are bullet points and not negative signs:
sin = opposite/hypotenuse
cos = adjacent / hypotenuse
tan = opposite / adjacent.
Also remember the special properties of the 30 - 60 - 90 triangle.
This tells us that side AB = 9√3

Combining this info:
sin (C) = 9√3 ÷ 18 = √3 ÷ 2
cos(B) = 9 √3 ÷ 18 = √3 ÷ 2
tan(C) = 9 √3 ÷ 9 = √3
sin(B) = 9 ÷ 18 = 1/2
tan(B) = 9 ÷ 9√3 = 1 / √3 = √(3) ÷ 3

5 0

4 years ago

Define the variable plz ASAP!!!!

Ksivusya [100]

There are 20 questions on the quiz.

4 0

4 years ago

Maths trig functions please help

Nikitich [7]

1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$ .

We can confirm this algebraically.

$\sin(x) + 2 = -\cos(x) + 3$

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$ . For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$ .