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2 years ago

Can someone help me and solve the following equation showing work please? thank you!

Mathematics

1 answer:

dsp732 years ago

3 0

The angle made by an arc at the centre of the circle is double the angle made by the same arc at any point on the circle. The measure of the angle θ will be 45°.

<h3>What are an arc and subtended angle theorem?</h3>

According to the arc and subtended angle theorem, The angle made by an arc at the centre of the circle is double the angle made by the same arc at any point on the circle.

The angle made by an arc at the centre of the circle is double the angle made by the same arc at any point on the circle. Therefore, the measure of the angle θ will be half the measure of the ∠P.

As it can be observed the measure of the angle ∠P is 90°, therefore, the measure of the angle θ can be written as,

∠P = 2 × ∠θ

90° = 2 × ∠θ

∠θ = 45°

Hence, the measure of the angle θ will be 45°.

Learn more about Arc and Subtended angle Theorem:

brainly.com/question/1364009

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You might be interested in

Find the exact solution to the equation. 7-log2(x+5) = 6

Sindrei [870]

Answer:

x = -3

Step-by-step explanation:

You can try the answers to see which works. Or, you can actually solve the equation for x.

... 7 - log2(x+5) = 6

... 1 - log2(x+5) = 0 . . . . subtract 6

... 1 = log2(x+5) . . . . . . . add log2(x+5)

... 2 = x +5 . . . . . . . . . . . take the antilog

... -3 = x . . . . . . . . . . . . . subtract 5

5 0

3 years ago

Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.

morpeh [17]

Answer:

1) a) 0.0945

b) 0.1062

2) a) 0.0138

b) 0.00081

3) a) 0.00094

b) 0.00083

4) 301.68 cents

5) 0.0056

Step-by-step explanation:

Since there are 3 Red, 7 Green, and 10 Blue marbles.

Total number of marbles N = 20

Probability (Red) = 3/20

Probability (Green) = 7/20

Probability (Blue) = 10/20

1) the probability of picking 5 marbles and getting at least one red marble

A) with replacement

P(at least 1 red of 5)= (3/20 * 17/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 3/20 * 17/20 * 17/20)

P(at least 1 red of 5) = 0.0783 +0.0138 + 0.0024 = 0.0945

B) without replacement

P(at least 1 red of 5) = ( 3/20 * 17/19* 16/18 * 15/17 * 14/16) + (3/20 * 2/19 * 17/18 * 16/17 * 15/16) + (3/20 * 2/19 * 1/18 * 17/17 * 16/16)

P(at least 1 red of 5) = 0.0921 + 0.01316 + 0.0009 = 0.1062

2) the probability of picking 6 marbles having 2 of each color

A) with replacement

P( 6, 2 of each) = 3/20 * 3/20 * 7/20 * 7/20 * 10/20 * 10/20

P( 6, 2 of each) = 0.0138

B) without replacement

P( 6, 2 of each) = 3/20 * 2/19 * 7/18 * 6/17 * 10/16 * 9/15

P( 6, 2 of each) = 0.00081

3) Pick 8 marbles: 4 green and 4 blue

A) with replacement

P(8, 4G and 4 B) = 7/20*7/20*7/20*7/20*10/20*10/20*10/20

P(8, 4G and 4 B) = 0.00094

B) without replacement

P(8, 4G and 4 B) = 7/20*6/19*5/18*4/17*10/16*9/15*8/14*7/13 = 0.00083

4) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

Cost of 6 .marbles= 6* 50 cents

C = 300 cents

Therefore, You will have to pay

(1 + 0.0056) 300 cent = 301.68 cents to be sure of getting at least 6 marbles of the same color

5) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

6 0

3 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

3 years ago

under a dilation, the point (2,6) is moved to (6,18) what is the scale factor of the dilation? enter your answer in the box

Anestetic [448]

Step-by-step explanation:

The equation of line passing through the two points is $\frac{y-y_{1} }{x-x_{1} }=\frac{y_{2} -y_{1}}{x_{2} -x_{1}}$

When substituted,the equation becomes $\frac{y-6}{x-2}=\frac{18-6}{6-2}$

which when simplified is $y=3\times x$

The line clearly passes through origin.

The distance between two points is $\sqrt{(y_{1}-y_{2} )^{2}+(x_{1}- x_{2} )^{2}}$

Distance between origin and $(2,6)$ is $\sqrt{40}$ .

Distance between origin and $(6,18)$ is $\sqrt{360}$

Scale factor is $\frac{distance\text{ }of\text{ }p_{2}\text{ from origin}}{distance \text{ } of \text{ } p_{1}\text{ from origin}}$

So,scale factor is $\frac{\sqrt{360} }{\sqrt{40} }$

which when simplified becomes 3.

6 0

3 years ago

When you multiply and divide by a negative you have to _______

Molodets [167]

Answer:

Multiplying and dividing with integers. You also have to pay attention to the signs when you multiply and divide. There are two simple rules to remember: When you multiply a negative number by a positive number then the product is always negative.

Step-by-step explanation:

8 0

3 years ago