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2 years ago

15

Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approxi

mately 5.88*10*10*10*10*10*10*10*10*10*10*10*10 light-years from Earth. In scientific notation, approximately how many miles is it?

1 answer:

Blababa [14]2 years ago

4 0

The earth is about 2.5578*10'13 miles

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an aircraft carrier leads port heading due west at 28 knots. A helicopter is 175 nautical miles from the carrier at an angle of

emmainna [20.7K]

Let’s say it takes time t hours for the interception.
In that time the carrier travels 28t nm west and the helicopter 130t nm.
Now we use the sine rule to find the angle x which lies between the distance 175 and 130t.
sin(x)/28t=sin35/130t. So sin(x)=28sin35/130=14sin35/65=0.1235 and x=7.0964 degrees.
Therefore the helicopter should use a bearing of 35+7.0964=42.0964 degrees north of east or 47.9 degrees east of north approx.

8 0

3 years ago

Eliza started her savings account with $100. Each month she deposits $25 into her account. Determine the average rate of change

Afina-wow [57]

First, lets create a equation for our situation. Let $x$ be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span> $f(x)=25x+100$
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is: $\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1} }$ . So lets replace the values in our formula to find the average of our function:
$\frac{25(10)+100-[25(2)+100]}{10-2}$
$\frac{350-150}{8}$
$\frac{200}{8}$
$25$

We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>

6 0

2 years ago

pllllllls I need help right triangle and trigonometry special right triangles find the value of each variable

yanalaym [24]

$\huge {\mathfrak{Answer : }}$

we know,

$\boxed{\cos( \theta) = \dfrac{base}{hypotenuse}}$

so,

$\cos(30) = \dfrac{24}{x}$

$\dfrac{ \sqrt{3} }{2} = \dfrac{24}{x}$

$x = \dfrac{24 \times 2}{ \sqrt{3} }$

$x = 8 \sqrt{3} \times 2$

$x = 16 \sqrt{3}$

now,

$\boxed{ \tan( \theta) = \frac{perpendcular}{base} }$

$\tan(30) = \dfrac{y}{24}$

$\dfrac{1}{ \sqrt{3} } = \dfrac{y}{24}$

$y = \dfrac{24}{ \sqrt{3} }$

$y = 8 \sqrt{3}$

6 0

2 years ago

Help on this one please??

xenn [34]

Answer:

C = ~50 deg

Step-by-step explanation:

Apply the cosine theorem:

cos(C) = (CA^2 + CB^2 - AB^2)/(2*CA*CB)

= (7.5^2 + 6.5^2 - 6^2)/(2*7.5*6.5)

= 0.6410

=> C = ~50 deg

Hope this helps!

7 0

3 years ago

Fiano borrowed $550. Four months later, he sent a cheque for $562.83 to repay the loan and interest. What was the interest rate

Digiron [165]

Answer:

Rate = 0.7%

Step-by-step explanation:

Rate = Interest / (principal x time)

rate = 12.83 / ((550) x 12/4)

rate = 12.83 / (550 x 3)

rate = 12.83 / 1650

rate = 0.007

rate = 0.7%

4 0

2 years ago