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2 years ago

Hurry im running out of time

Mathematics

1 answer:

Dimas [21]2 years ago

3 0

Answer:

it is ∠EBA and ∠DBC i think

Step-by-step explanation:

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If pamela has 20 of string does she have enough to make bracelets for 4 of her friends?

JulijaS [17]

How much string is needed to make one bracelet?

From the information given, she should have enough to make 4 bracelets for each of her friends if she uses 5 strings per bracelet.

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3 years ago

Can someone please check my answers I WIILL GOVE BRAINLY TO FIRST PERSON THAT ANSWERS!

Rainbow [258]

The first is C. r = 4(4/9)
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3 years ago

A (0,5)<br> B (3, 11)<br> What’s the equation

slava [35]

Y=2x+5
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6 0

1 year ago

A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate

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

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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