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steposvetlana [31]

3 years ago

6

Will upvote brainiest answer and add friend

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

68°

Step-by-step explanation:

ABD is a right angle and the measure of it is 90°

m<CBD + m<ABC = m<ABD

4x + 52° + 8x - 10 = 90° add like terms

12x + 42° = 90°

12x = 48°

x = 4

m<CBD is 4x + 52° replace x with 4

4×4 + 52 = 68°

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A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determi

grandymaker [24]

The mass of the first shipment at time t is
$m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}$

The mass of the second shipment at time t is
$m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }$

At time t, the ratio of m₁ to m₂ is
$\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123$

Therefore as a percentage,
$\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%$

Answer: B. 81.2%

4 0

3 years ago

NEED HELP SOON!!!!<br> for any numbers a, b, and c, a(b + c)=ab+ac

navik [9.2K]

The property displayed here is the distributive property.

If you have a variable or unknown number inside or outside of parentheses, you can distribute it to each term and add the terms together, and it will remain true.

Example:

4(x + 5)

After distributing, it'll look like this:

4x + 20

6 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

2 years ago

HELP!!! What is the slant height x of the square pyramid? Express your answer in radical form.

katrin [286]

Check the picture below.

make sure your calculator is in Degree mode.

3 0

3 years ago

When QS=6x and CS=5x-8, what is AB?

patriot [66]

Answer:

AB = x+ 8

Step-by-step explanation:

Given : QS=6x and CS=5x-8

From the given figure,

QS = QC + CS

We are given with QS and CS

Replace it the above equation

6x = QC + 5x -8

Solve for QC

Subtract 5x on both sides

x = QC + -8

Add 8 on both sodes

QC= x + 8

QABC is a rectangle

QC = AB

So AB = x+ 8

4 0

3 years ago