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3 years ago

Solve for x in the diagram below. x 3xº 100°

Mathematics

2 answers:

Harrizon [31]3 years ago

7 0

Answer:

x+3x+100°=180°(straight line is 180°)

4x+100°=180

4x=180°-100°

4x=80

x=80/4

x=20

diamong [38]3 years ago

6 0

Ummm ....... I think x=80

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An object is launched from a platform. It's height (in meters), x seconds after the launch, is modeled by: h(x)=-5(x-4)^2+180. W

Kipish [7]

The height of the object at the time of launch is 100 meters.

To find this, we simply have to put 0 in for x, as this is when there has been no time (at launch)

h(x)=-5(x-4)^2+180

h(0)=-5(0-4)^2+180

h(0)=-5(-4)^2+180

h(0)=-5(16)+180

h(0)=-80+180

h(0) = 100

7 0

4 years ago

Given the points R (6,-2) and T (-9,- 7), find the coordinates of point S on line RT such that the ratio of RS to ST is 3 : 2 p

Sholpan [36]

Answer:

(-3, -5)

Step-by-step explanation:

By using the distance formula, we see that the length of RT is $\sqrt{(6+9)^2+(-2+7)^2}$ which equals $5\sqrt{10}$ .

The ratio of RS to ST is 3:2, so we can write the distance of RS as $(5\sqrt{10} / 5) * 3$ which equals $3\sqrt{10}$ . The distance of ST is then $2\sqrt{10}$ .

Since we know the distance of RS and ST, we can find the coordinates of point S by using the distance formula again:

$\sqrt{(6-x)^2+(-2-y)^2} = 3\sqrt{10}$ and $\sqrt{(-9-x)^2+(-7-y)^2} = 2\sqrt{10}$

We solve for x and y, getting (-3, -5).

3 0

3 years ago

Determine the value of the angle.<br><br><br><br> Angle BDC=

Nadusha1986 [10]

90°-53°=37°

The whole angle is right angled

7 0

2 years ago

Read 2 more answers

A girl's club held a fundraiser last year and raised $600 for the local soup kitchen. This year they raised $960. What is

Gnesinka [82]

Answer:

60%

Step-by-step explanation:

5 0

3 years ago

Can you find the probability to each of these answers...

goldfiish [28.3K]

Problem 1

<h3>Answer: 7/10</h3>

-----------------------

Explanation:

The formula we'll use is

P(A or B) = P(A) + P(B)

which only works if A and B are mutually exclusive events.

P(A or B) = P(A) + P(B)

P(A or B) = 7/20 + 7/20

P(A or B) = (7+7)/20

P(A or B) = 14/20

P(A or B) = (7*2)/(10*2)

P(A or B) = 7/10

=========================================================

Problem 2

<h3>Answer: 3/4</h3>

-----------------------

Explanation:

We'll use the same formula as the previous problem.

P(A or B) = P(A) + P(B)

P(A or B) = 3/10 + 9/20

P(A or B) = 6/20 + 9/20

P(A or B) = (6+9)/20

P(A or B) = 15/20

P(A or B) = (3*5)/(4*5)

P(A or B) = 3/4

=========================================================

Problem 3

<h3>Answer: 3/5</h3>

-----------------------

Explanation:

We'll use the same formula as the previous problem.

P(A or B) = P(A) + P(B)

P(A or B) = 7/20 + 1/4

P(A or B) = 7/20 + 5/20

P(A or B) = (7+5)/20

P(A or B) = 12/20

P(A or B) = (4*3)/(4*5)

P(A or B) = 3/5

=========================================================

Problem 4

<h3>Answer: 0</h3>

-----------------------

Explanation:

This time we're asked to find P(A and B), but since the two events are mutually exclusive, this means the probability of both occurring is 0.

Mutually exclusive events cannot happen simultaneously.

An example would be flipping heads and tails at the same time on the same coin.

The info about P(A) and P(B) is not relevant.

7 0

4 years ago