All categories

2 years ago

A triangle was dilated by a scale factor of 2. If cos a° = three fifths and segment FD measures 6 units, how long is segment DE?

Mathematics

2 answers:

kiruha [24]2 years ago

6 0

Answer:

10 units

Step-by-step explanation:

Given that:

Cos a = 3 /5

From trigonometry ;

Cos θ = Adjacent / hypotenus

Side Adjacent angle a = 3 = FD

Hypotenus = 5 = DE

3 / 5 = FD / DE

FD = 6 units

3/5 = 6/DE

DE * 3/5 = 6

DE = 6 ÷ 3/5

DE = 6 * 5 /3

DE = 30 /3

DE = 10 units

Artist 52 [7]2 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

What is y+4=9 what is y?

amid [387]

Answer:

$y = 5$

Step-by-step explanation:

Subtract the 4 from 9

5 0

3 years ago

Read 2 more answers

24 red marbles is40% of ? Marbles

Anvisha [2.4K]

Answer: The answer is 60

Step-by-step explanation: 40% = .4 because % means hundredths. So 40% = 40/100 = .4

Let x be the total number of marbles.

Then, .4x = 24

Divide both sides by .4

x = 60 and 24 is 40% of 60 marbles

5 0

3 years ago

Read 2 more answers

How do I fine the sun and difference of number 16?

Julli [10]

Difference=subtract sum=add. just answering so you don't waste points :-)

6 0

3 years ago

What is the inverse of f(x) = sqft x+ 3

grin007 [14]

Answer:

$f^{-1} (x)=x^{2} -3$

Step-by-step explanation:

Given the function

$f\left(x\right)=\sqrt{x+3}$

We can find the inverse function by exchanging x and y, and then solve for y.

$y=\sqrt{x+3}$

Replace x with y.

$x=\sqrt{y+3}$

square both sides

$x^2=y+3$

$y=x^2-3$

Hence,

$f^{-1} (x)=x^{2} -3$

8 0

2 years ago

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

2 years ago