Just need an answer ASAP to pass this class.

Mathematics

2 answers:

svlad2 [7]3 years ago

8 0

X=11. Bisects means the angles on both sides of the line are equal, so 4x-16=2x+16 then solve for x

DedPeter [7]3 years ago

8 0

4x-16= 2x+6

4x-16+16= 2x+6+16

4x= 2x+22

4x-2x=22

2x=22

Divide by 2

2x= 22

x=11

Answer : second choice -11

You might be interested in

What is 0.000000591 as a scientific notation

Fiesta28 [93]

Answer:

5.91× 10^ −7

Step-by-step explanation:

Hope this helps

4 0

3 years ago

The function f(h)=m(1/2)^h gives the mass, m, of a radioactive substance remaining after h half-lives. Cobalt-60 has a half-life

goldfiish [28.3K]

Answer:

Third choice:

$f(x)=50(0.877)^{10};13.5mg$

Explanation:

1. Data:

Function:

$f(h)=m(1/2)^h$

Cobalt-60's half-life: 5.3 years

Initial mass of cobalt-60: 50 mg

2. Unknown:

equation for the mass of cobalt-60 remaining after 10 years = ?

mass remaining =?

3. Solution

Half-life is the time it takes a sample to decay to half of its initial amount. It is considered constant. Hence, when one half-life passes, the sample has decayed to 50% of the original amount; when two half-lives pass, the sample has decayed to (1/2)×(1/2) = 1/4 = 25%; when three half-lives have elapsed, the sample has decayed to (1/2)³ = 1/8 = 12.5% of its original amount, and so on.

Then, the amound of a sample remaining is calculated as the original amount times (1/2) raised to the number of half-lives elapsed, which is what the given function, $f(h)=m(1/2)^h$ models.

You just must substitute the data into the function to get the answer to the question:

$f(x)=50(0.5)^{(10/5.3)}$

Where, 50 is the original mass of 50g, 0.5 is equal to 1/2, and 10/5.3 gives the number of half-lives (the number of times that 5.3 years is contained in 10 years).

Simplifying:

$f(x)=50(0.5)^{(10/5.3)}=50((0.5)^{(1/5.3})^{10}\approx50(0.877)^{10}$