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3 years ago

14

An observer spots a plane flying at a 42 degree angle to his horizontal line of sight. If the plane is flying at an altitude of

15,000 ft, what is the distance (x) from the plane (P) to the observer (O)

1 answer:

Natalija [7]3 years ago

6 0

check the picture below.

make sure your calculator is in Degree mode.

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(4,8);m=3/4 pls help me

Crazy boy [7]

Answer:

y=3/4x+5

Step-by-step explanation:

4 0

3 years ago

A total of 234 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

lorasvet [3.4K]

234 = a + s s = students a = adults
2a = s
after you get the equations you solve. plug 2a into the above equation.
so u get
234 = a + 2a
234=3a then divide 3 from each side
78 = a
so 78 adult tickets were sold

3 0

3 years ago

Pls pls help me please!!!!

BigorU [14]

Answer:

Move 6.2 units up

Step-by-step explanation:

(-5.4, 6.2)

The first coordinate is the x coordinate

Positive x moves to the right, negative x moves to the left

Move 5.4 units to the left

The second coordinate is the y coordinate

Positive y moves to up, negative y moves down

Move 6.2 units up

4 0

3 years ago

If a package of corn chips weighs 1.2 ounces what is the weight in ounces of 48 packages ?

Naya [18.7K]

I'm not usually good at these questions but you can get a calculator that does weight multiplied by units.
so with your question it would be :

1.2 ounces x (48 units)
= 57.6 ounces

57.6 ounces

5 0

3 years ago

Read 2 more answers

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

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