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3 years ago

Find the slope in (5,1) and (-2,1)

Mathematics

2 answers:

xeze [42]3 years ago

8 0

Use the slope formula: $\frac{y_2-y_1}{x_2-x_1}$ . Substitute the appropriate coordinates into the formula.

y_2 = 1

y_1 = 1

x_2 = -2

x_1 = 5

$\frac{1-1}{-2-5}$ = $\frac{0}{-7} =0$ . The slope is 0.

Elden [556K]3 years ago

5 0

Use the slope formula

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Curtis wants to save money for a trip overseas. Curtis invests $900 in an account that pays an interest rate of 8.75%. How many

professor190 [17]

Hi there
The formula is
A=p (1+r)^t
A future value 11700
P present value 900
R interest rate 0.0875
T time?
We need to solve for t
T=log (A/p)÷log (1+r)
So
T=log(11,700÷900)÷log(1+0.0875)
T=30.6years round your answer to get
T=31 years

Hope it helps

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3 years ago

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The population of an island is 6000 people this year and is projected to be 6300 people next year, by what percentage is the pop

Julli [10]

Answer:

5% increase

Step-by-step explanation:

First, find how much it increased:

6300 - 6000

= 300

Then, to find the percent increase, divide 300 by 6000

300/6000

= 0.05

So, there was a 5% increase in the population

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3 years ago

If a day was a year, how many does it take to make a million years?

Tasya [4]

It would take 2,740 days to make a million years.

1,000,000 divided by 365.

6 0

3 years ago

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A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

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2 years ago

The polynomial below is perfect square trinomial of the form a^2 -2AB +B^2

guajiro [1.7K]

Answer:

In order to get this perfect square trinomial, the factored form will be (a - b)^2

Step-by-step explanation:

We know this because it follows the rule of perfect squares in which the ends are both squares and the middle term is -2 times each term.

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3 years ago