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3 years ago

Ughhh this is hard for me!

Mathematics

1 answer:

Annette [7]3 years ago

4 0

Answer:

(x+4)/3. When x is 5 the answer is 3

Step-by-step explanation:

You might be interested in

Sam bought a new couch on sale at the discount furniture store. The original cost of the couch was $550. The sale price was $449

dolphi86 [110]

30%x550=165
550-165=385
385+99=385.00

4 0

2 years ago

Can you bisect an angle using only a compass and a straight edge

ahrayia [7]

Answer:

yes

Step-by-step explanation:

bisect = dividing it into two equal parts

if you were to line the angle up with the north (or really any line of the compass), you could find the center (you could find where two lines [the same distance from north] are on equal sides of north, or you could just find the center of the angle from sight)

and with the straight edge, you could easily divide the angle into two equal halves

4 0

2 years ago

Quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm. Which could be the side lengths of a dilation of HIJK wit

matrenka [14]

Answer:

Step-by-step explanation:

The quadrilateral HIJK has sides measuring 12 cm, 26 cm, 14 cm, and 30 cm.

When HIJK is dilated with a scale factor of 1.5, the side lengths becomes:

12 X 1.5 =18 cm

26 X 1.5 =39 cm

14 X 1.5 =21 cm

30 X 1.5 =45 cm

A dilation of HIJK with a scale factor of 1.5 will give us the side lengths:

18cm, 39cm, 21cm and 45cm.

<u>The correct option is C.</u>

6 0

3 years ago

Solve the word problem. The table shows the low outside temperatures for Monday, Tuesday, and Wednesday. What was the change in

den301095 [7]

Answer: D. -3.2

Step-by-step explanation:

The correct answer to this question is D. -3.2 because when you subtract positive 8.8 and positive 12 your answer is -3.2.

(+8.8) - (+12) = -3.2

7 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

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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