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3 years ago

Harold and his friend Bill each collect vintage action figures. Harold already has

Mathematics

1 answer:

Neporo4naja [7]3 years ago

5 0

Harold: y = 3x + 18

Bill: y = 2x+24

Hope this helps with your mid-term too guys !

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Garrett can complete four laps around the track in 7 minutes and 45 seconds. How many FULL LAPS can he complete in 1 hour?

tiny-mole [99]

Answer: N/A

Step-by-step explanation:

this answer is not physically possible cause the human body is not capable of going that fast continually for an hour. If you are in athletics and run everyday you would understand. i mean come on don't believe me? your try.

4 0

3 years ago

2+2=4 then what is 4+4 ps. just get some brainly points

FinnZ [79.3K]

Answer:

Step-by-step explanation:

because yes.

5 0

3 years ago

Read 2 more answers

Y=2x-4 What is the five ordered pairs and completing the input/output table

kati45 [8]

Answer:

{(0, -4), (1, -2), (2, 0), (3, 2), (4, 4)}

Input values (x) = {0, 1, 2, 3, 4}

Output values (y) = {-4, -2, 0, 2, 4}

Step-by-step explanation:

Given the linear equation, y = 2x - 4, you could practically choose any input value (for x), to have an output value (for y).

If x = 0:

y = 2(0) - 4

y = 0 - 4

y = -4

First ordered pair: y-intercept = (0, -4)

If x = 1:

y = 2(1) - 4

y= 2 - 4

y = -2

Second ordered pair: (1, -2).

If x = 2:

y = 2(2) - 4

y= 4 - 4

y = 0

Third ordered pair: x-intercept = (2, 0).

If x = 3:

y = 2(3) - 4

y = 6 - 4

y = 2

Fourth ordered pair: (3, 2).

If x = 4:

y = 2(4) - 4

y = 8 - 4

y = 4

Fifth ordered pair: (4, 4).

Your input values (x) = {0, 1, 2, 3, 4}

Your output values (y) = {-4, -2, 0, 2, 4}

6 0

3 years ago

Solve the system of equations. y=23x–19 x2–y= – 6x–23 Write the coordinates as integers, simplified proper or improper fractions

forsale [732]

Answer:

(3, 50) and (14,303)

Step-by-step explanation:

Given the system of equations;

y=23x–19 ....1

x²–y= – 6x–23 ...2

Substitute 1 into 2;

x²–(23x-19)= – 6x–23

x²–23x+19= – 6x–23 .

x²-23x + 6x + 19 + 23 = 0

x² - 17x + 42 = 0

Factorize;

x² - 14x - 3x + 42 = 0

x(x-14)-3(x-14) = 0

(x-3)(x-14) = 0

x = 3 and 14

If x = 3

y = 23(3) - 19

y = 69-19

y = 50

If x = 14

y = 23(14) - 19

y = 322-19

y = 303

Hence the coordinate solutions are (3, 50) and (14,303)

8 0

3 years ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

——————————

Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

3 years ago