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4 years ago

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When solving for x, the sine of 33º is used. What other angle and trigonometric function could be used to solve for x? a. cos 57

º c. tan 57º b. tan 33º d. cos 33º
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1 answer:

Morgarella [4.7K]4 years ago

5 0

Answer:

The answer is A.Cos 57.

Step-by-step explanation:

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Lunna [17]

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3 years ago

01.01) Rewrite the expression with rational exponents as a radical expression. 7 times x to the two thirds power cube root of th

jeka94

Answer:

$(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2$

Step-by-step explanation:

Given the expression $(7x)^{\frac{2}{3}$ , to express this as a radical expressions, we'd apply the rule/law of indices that deals with converting expressions that has rational exponents into radical expressions.

The rule of indices to apply is: $b^{\frac{m}{n}} = (\sqrt[n]{b})^m$

To apply this to the expression, $(7x)^{\frac{2}{3}$ , the denominator of the fraction of the exponent would determine the root, that is, cube root in this case. The numerator of the exponent would then determine the exponent of the radical expressions.

Thus:

$(7x)^{\frac{2}{3} = (\sqrt[3]{7x})^2$

6 0

3 years ago

If an arrow is shot upward on Mars with a speed of 62 m/s, its height in meters t seconds later is given by y = 62t − 1.86t². (R

Soloha48 [4]

Answer:

Approximately $58.28\; \rm m \cdot s^{-1}$ .

Step-by-step explanation:

The velocity of an object is the rate at which its position changes. In other words, the velocity of an object is equal to the first derivative of its position, with respect to time.

Note that the arrow here is launched upwards. (Assume that the effect of wind on Mars is negligible.) There would be motion in the horizontal direction. The horizontal position of this arrow will stays the same. On the other hand, the vertical position of this arrow is the same as its height: $y = 62\, t - 1.86\, t^2$ .

Apply the power rule to find the first derivative of this $y$ with respect to time $t$ .

By the power rule:

the first derivative of $t$ (same as
the first derivative of $t^2$ (same as $t$ to the second power) with respect to

Therefore:

$\begin{aligned}\frac{dy}{d t} &= \frac{d}{d t}\left[62 \, t - 1.86\, t^2\right] \\ &= 62\,\left(\frac{d}{d t}\left[t\right]\right) - 1.86\, \left(\frac{d}{d t}\left[t^2\right]\right) \\ &= 62 \times 1 - 1.86\times\left(2\, t) = 62 - 3.72\, t\end{aligned}$ .

In other words, the (vertical) velocity of this arrow at time $t$ would be $(62 - 3.72\, t)$ meters per second.

Evaluate this expression for $t = 1$ to find the (vertical) velocity of this arrow at that moment: $62 - 3.72 \times 1 =58.28$ .

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3 years ago

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Lauren bought 6 yellow roses, 10 orange roses, and 12 pink roses to make a bouquet. What is the ratio of the number of yellow ro

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3 years ago