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2 years ago

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[Brainliest Answer] A ball's position, in meters, as it travels every second is represented by the position function s(t)=4.9t^2

+450.
What is the velocity of the ball after 5 seconds?
Include units in your answer.

1 answer:

Vikentia [17]2 years ago

7 0

Should be right
$2200$

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The angle \theta_1θ

enyata [817]

Answer:

$sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Step-by-step explanation:

It is given that:

$cos\theta_1 = -\dfrac{12}{19}$

And we have to find the value of $sin\theta_1 = ?$

As per trigonometric identities, the relation between $sin\theta\ and \ cos\theta$ can represented as:

$sin^2\theta + cos^2\theta = 1$

Putting $\theta_1$ in place of $\theta$ Because we are given

$sin^2\theta_1 + cos^2\theta_1 = 1$

Putting value of cosine:

$cos\theta_1 = -\dfrac{12}{19}$

$sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}$

It is given that $\theta_1$ is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

$\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}$

8 0

2 years ago

In the diagram, the radius of the outer circle is 2x cm and the radius of the inside circle is 6 cm. The area of the shaded regi

Minchanka [31]

The picture in the attached figure

we know that
Area of circle=pi*r²
area of the shaded region=364*pi cm²
area of the shaded region=area of the outer circle -area of the<span> inside circle

</span>area of the outer circle=pi*[2x]²----> 4*x²*pi cm²
area of the inside circle=pi*6²------> 36*pi cm²
364*pi=4*x²*pi-36*pi------> 364=4*x²-36
4*x²=364+36-----> 4*x²=400
x²=400/4----> x²=100
x=10 cm

the answer is
x=10 cm

4 0

2 years ago

∆ABC is translated 2 units down and 1 unit to the left. Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

RideAnS [48]

Answer:

The coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

Step-by-step explanation:

From the figure it is clear that the coordinates of triangle are A(0,0), B(1,3) and C(1,1).

∆ABC is translated 2 units down and 1 unit to the left.

$(x,y)\rightarrow (x-1,y-2)$

$A(0,0)\rightarrow (0-1,0-2)=(-1,-2)$

$B(1,3)\rightarrow (1-1,3-2)=(0,1)$

$C(1,1)\rightarrow (1-1,1-2)=(0,-1)$

Then it is rotated 90° clockwise about the origin to form ∆A′B′C′.

$(x,y)\rightarrow (y,-x)$

$(-1,-2)\rightarrow A'(-2,1)$

$(0,1)\rightarrow B'(1,0)$

$(0,-1)\rightarrow C'(-1,0)$

Therefore the coordinates of image are A'(-2,1), B'(1,0) and C'(-1,0).

7 0

1 year ago

Problem Lois and Clark own a company that sells wagons. The amount they pay each of their sales employees (in dollars) is given

Oxana [17]

12h + 30w.....where h = hrs worked and w = wagons sold

so if an employee works 6 hrs and sells 3 wagons....then h = 6 and w = 3

12h + 30w
12(6) + 30(3) =
72 + 90 = $ 162 <==

8 0

2 years ago

There are 345 students at a college who have taken a course in calculus, 212 who have taken a course in discrete mathematics, an

ollegr [7]

Answer:

369 students have taken a course in either calculus or discrete mathematics

Step-by-step explanation:

I am going to build the Venn's diagram of these values.

I am going to say that:

A is the number of students who have taken a course in calculus.

B is the number of students who have taken a course in discrete mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the number of students who have taken a course in calculus but not in discrete mathematics and $A \cap B$ is the number of students who have taken a course in both calculus and discrete mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

188 who have taken courses in both calculus and discrete mathematics.

This means that $A \cap B = 188$

212 who have taken a course in discrete mathematics

This means that $B = 212$

345 students at a college who have taken a course in calculus

This means that $A = 345$

How many students have taken a course in either calculus or discrete mathematics

$(A \cup B) = A + B - (A \cap B) = 345 + 212 - 188 = 369$

369 students have taken a course in either calculus or discrete mathematics

4 0

2 years ago