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3 years ago

The United States Air Force has 19 women for every 81 men in enlisted if a squadron has 750 members how many women are there

Mathematics

1 answer:

elena-s [515]3 years ago

3 0

Answer:

the answer is B

Step-by-step explanation:

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A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 27 mi/hr. Another ship leaves the same port at 1:30 P.M. and tr

Len [333]

To solve this problem you must apply the proccedure shown below:

1. You must apply the Law of Cosines, as you can see in the figure attached. Then:

- The first ship travels at $27 mi/hr$ in for two hours. Therefore, the side $a$ is:

$a=(27 mi/hr)(2 hr)=54mi$

- The second ship travels at $18 mi/hr$ for $1.5 hours$ . Therefore, the side $b$ is:

$b=(18mi/hr)(1.5hr)=27mi$

- Now, you can calculate $c$ :

$c=\sqrt{54^{2}+27^{2}-2(54)(27)Cos(55)}=44 mi$

The answer is: $44 miles$

8 0

3 years ago

If 3a 2b=24 and 4a 5b=53, what is the value of a b?

Alexus [3.1K]

7a+7b+77
I I hope this helps you

4 0

3 years ago

What is 1/2 y − 5 =−7

alexgriva [62]

Answer:

=-4

Step-by-step explanation:

4 0

3 years ago

Solve for y x=y^2+4y

Lorico [155]

$x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\$

-----------------------------------------------------

$x=y^2+4y\\ y^2+4y-x=0\\
y^2+4y+4-4-x=0\\
(y+2)^2=x+4\\
y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\$

4 0

3 years ago

Use the photo at the bottom:

satela [25.4K]

Answer:

<2 = 129°

Step-by-step explanation:

if I || m then 5x+12 =m<2 and m<2 + 2x + 14 = 180°

2x + 14 + 5x + 12 = 7x + 26 =180°

7x = 154 ➡x = 21°

m <2 = 5×21 + 14

8 0

3 years ago