a ship is 45 miles east and 30 miles south of port. if the captain wants to sail directly to south port, what beating sold be taken

Answers

Answer 1

Answer: probaly you meant 'what bearing should be taken'

so use pythagorean theorem
c=distance from position to south port
b=45
a=30

a^2=B^2=c^2
30^2+45^2=c^2
900+2025=c^2
2925=c^2
square root of both sides
15 times √13
aprox
54.08
54.1
opposite of east=west
opposite of south=north

bearing=
54.1 miles north-west

Answer 2

Answer: Well
  East
45 miles > > > >
< v
< v 30 miles South
North < v
West   < Δ= ship

A right angle is formed, so we use Pythagoras.

$45 ^(2) +30 ^(2)=2925$
$√(2925)=54.08$

Therefore,  the route should be 54 miles North West

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A 26‘long painting is how many yards long?

Answers

Hello!

If a painting is 26 feet long, it's 8 and 2/3 yards long.

This is because there is 3 feet in a yard. You would do 26 divided by 3. You would get 8 yards and have 2 feet left. Obviously 2 feet is smaller than a yard, so it'd be 2/3 of a yard. Therefore the answer is 8 and 2/3 yards.

Hope this helped! :)

Which pair of rectangles are similar

Answers

Answer:

Its 2nd one

Step-by-step explanation:

189÷90=2.1

81÷90=0.9

The expression sinx(cscx - cotx cosx) can be simplified to.

Answers

csc x = 1 / sin x and cot x = cos x/ sin x ;
sin x * ( csc x - cot x * cos x ) =
= sin x * ( 1/sin x - cos x / sin x * cos x ) =
= sin x * ( ( 1 - cos² x ) / sin x ) =
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My grandparents have four grandchildren. The product of the ages of the four grandchildren is 67 184. The youngest grandchild is younger than 10, and is also 30 years younger than the oldest grandchild.Determine all possibilities for the ages of my grandparents’ grandchildren.

Answers

Answer:

The possible ages of the four grandchildren are a = 4, b = 19, c = 26, and d = 34

Step-by-step explanation:

The given parameters are;

The number of grandchildren in the family = 4

The product of the ages of the four grand children = 67184

The age of the youngest grandchild < 10

The age of the oldest grandchild = 30 + The age of the youngest grandchild

Let a represent the age of the youngest grandchild, and let b, and c represent the ages of the other two intermediate grandchild

Therefore, we have;

a < 10

The age of the oldest grandchild = a + 30 < 10 + 30

∴ The age of the oldest grandchild < 40

The product of the ages of the four grandchildren = a × b × c × (a + 30) = 67184

The factors of 67184 that are between 1 and 40 are;

1, 2, 4, 8, 13, 16, 17, 19, 26, 34, 38

Taking a = 8, we have;

The age of the oldest grandchild × The age of the youngest grandchild = a × (a + 30) = 8 × 38 = 342

Therefore. a × b = 67184/(a × (a + 30) = 196.44

Therefore, a ≠ 8

For a = 4, we have the age of the oldest grandchild = a + 30 = 4 + 30 = 34

The age of the oldest grandchild × The age of the youngest grandchild = a × (a + 30) = 4 × 34 = 136

Therefore. a × b = 67184/(a × (a + 30) = 494

We find that the other factors of 67184, which are 19 and 26 have a product of 494

Therefore, the possible ages of the four grandchildren are a = 4, b = 19, c = 26, and d = 34

To give, 4 × 19 × 26 × 34 = 67,184.

The volume of cylinder is 448 pie cm cube and height 7 cm find its radius

Answers

Answer:

$r =\bf 8 \space\ cm$

Step-by-step explanation:

The formula for volume of a cylinder is as follows:

$\boxed{Volume = \pi r^2 h}$

where:

• r = radius (? cm)

• h = height (7 cm).

Substituting the values into the formula:
$448 \pi = \pi * r^2 * 7$

Now solve for $r$ :

⇒ $r^2 = (448 \pi )/(\pi * 7)$

⇒ $r = √(64)$

⇒ $r =\bf 8 \space\ cm$

$\huge\mathbb{ \underline{SOLUTION :}}$

Given:

$\longrightarrow\bold{Volume= 449 \pi cm^3}$

$\longrightarrow\bold{Height= 7cm}$

$\longrightarrow\sf{V= \pi r^2 h}$

$\longrightarrow\sf{448 \pi= \pi r^2 \cdot 7}$

$\longrightarrow\sf{448=</p><p>7r^2}$

$\longrightarrow\sf{r^2= (448)/(7) }$

$\longrightarrow\sf{r^2=64}$

$\huge \mathbb{ \underline{ANSWER:}}$

$\longrightarrow\sf{r= 8cm}$

If log 6=a, then log 600= ?

Answers

$\log 600 = \log (100 \cdot 6)= \log 100 + \log 6 = \log 10^2 + \log 6 = \n 2 \log 10 +\log 6 =2+\log 6 = \boxed{a+2} \n \hbox{From formulas:} \n \log a + \log b = \log (ab) \n \log a^b = b\log a$

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