All categories

3 years ago

How would you write the Pythagorean theorem for a right triangle with sides of 9 inches, 41 inches, and 40 inches?

Mathematics

1 answer:

VMariaS [17]3 years ago

5 0

A
9^2+21^3=80

B
9^2+50^3=100

C
60^2+20^2=80

You might be interested in

In which quadrant angle k lies, if tan k<0, an csc k >

Zanzabum

Answer:

A. QII.

Step-by-step explanation:

The tan < 0 in Quadrants 2 and 3.

The csc > 0 in Quadrants 1 and 2.

7 0

3 years ago

Which graph shows the solution of the inequality −|x − 2| + 9 > 6?

mariarad [96]

answer is -1 < x < 5

-|x - 2|+ 9 > 6

Rearrange the terms

-|x - 2| > 6 - 9

-|x - 2| > - 3

then divide both sides of the inequality by the co- efficient of variable

|x - 2| < 3

convert the absolute inequality to standard inequality

-3 <x - 2 < 3

separate compound inequalities into system of inequality

{x - 2}> -3

{x - 2 < 3}

Rearrange variable to the left side of the equation

x > -3 + 2

calculate the sum or difference

x > -1

x -2 < 3

Rearrange variable to the left side of the equation

x < 3 + 2

calculate the sum or difference

x < 5

x > -1 and x < 5

Find intersection

-1 < x < 5

Equations and inequalities are both mathematical sentences formed by relating two expressions to each other. In an equation, the two expressions are deemed equal which is shown by the symbol =. Where as in an inequality, the two expressions are not necessarily equal which is indicated by the symbols: >, <, ≤ or ≥.

learn more about inequality equation here :https://brainly.in/question/15934172

SPJ9

4 0

1 year ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$