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2 years ago

11

An airplane began its descent when it was 0.6 mile above the ground and 15 miles away from an airport, as shown in the diagram b

elow.

1 answer:

mart [117]2 years ago

5 0

Answer:

sin⁻¹(0.6/15)

Step-by-step explanation:

The equation for sine is Y = A sin Bx

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The two roots a+sqrb and a-sqr b are called _______radicals.

astraxan [27]

Answer:

Conjugate radicals.

Step-by-step explanation:

The two roots a+sqrb and a-sqr b are called conjugate radicals.

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2 years ago

Evaluate the determinant for the following matrix [3 -5 / 1 1]

Natalka [10]

Answer:

<h2>A. -2</h2>

Step-by-step explanation:

$\det\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] =ad-bc\\\\\det\left[\begin{array}{ccc}3&-5\\1&1\end{array}\right] =(3)(1)+(-5)(1)=3-5=-2$

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2 years ago

Help QUICKLY, will offer brainliest for correct answer

FromTheMoon [43]

Answer:

48

Step-by-step explanation:

-2(1)^2(3(1)^2 – 7(1) + 10)

-2^2(3^2 – 7 + 10)

4(9 - 7 + 10)

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48

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3 years ago

Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the

docker41 [41]

Answer:

15 miles

Step-by-step explanation:

Let $x$ be the miles in the circular park path, $t_{L}$ the time Louisa takes to finish and $t_{C}$ the time Calvin takes to finish both in hours.

Then $x$ , the longitude is equal to the velocity times the time used to finish. So

$x=5t_{L}$

$x=6t_{C}$

And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:

$t_{C}=t_{L}-0.5$

Three equations, three unknowns, the system can be solved.

Equalizing the equation with x :

$5t_{L}=6t_{C}$

In this last equation replace $t_{C}$ with the other equation and solve:

$5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3$

With Louisa's time find x:

$x=5t_{L}\\ x=5(3)\\ x=15$

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2 years ago

-1\dfrac{60}{100} + 0.1 + \dfrac{1}{4} =−1 100 60 +0.1+ 4 1

Jlenok [28]

Answer: Your question doesn't make any sense

Step-by-step explanation:

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3 years ago