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4 years ago

HELP GUYS IM REALLY NOT SURE,,

Mathematics

1 answer:

Solnce55 [7]4 years ago

5 0

What u not sure bout?

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I NEED HELPPP<br><br> Whats the length of LN?

andre [41]

Answer:

LN = 66 cm

Step-by-step explanation:

The 3 parallel lines divide the 2 sides they intersect in proportion, that is

$\frac{DE}{EF}$ = $\frac{LM}{MN}$ , substitute values

$\frac{7}{15}$ = $\frac{LM}{45}$ ( cross- multiply )

15LM = 315 ( divide both sides by 15 )

LM = 21

Hence

LN = LM + MN = 21 + 45 = 66 cm

3 0

3 years ago

Using the numbers 5, 8, and 24, create a problem using no more than 4 operations (adding, subtracting, multiplication, division,

nalin [4]

<span>√(5*8*24) or anything along those lines, you can switch the numbers around , it wont matter</span>

3 0

3 years ago

You are making juice from concentrate. The directions on the packaging say to mix

mel-nik [20]

Answer:

25 can.

Step-by-step explanation:

Here is the complete question: You are making juice from concentrate. The directions on the packaging say to mix 1 can of juice with 3 cans of water to make orange juice. How many 12 fluid ounces cans of the concentrate are required to prepare 200 6-ounce servings of orange juice?

Given: Ratio to make juice with concentrate:water is 1:3.

As required we need to prepare 200 6 ounce serving of Orange juice.

∴ $200\times 6= 1200\ ounce$

Let there be x ounce of concentrate and 3x ounce of water to make 1200 ounce of orange juice.

Now, $x+3x= 1200$

∴ x= 300 ounce

Next, lets find out how many cans of 12 ounce is required.

$\frac{300}{12} = 25\ cans$

∴ 25 cans is required to make 1200 ounce of orange juice.

3 0

3 years ago

If h(x) is the inverse of f(x). what is the value of h(f(x))?

attashe74 [19]

Option C is correct

The value of h(f(x)) is, x

Inverse function:

An inverse function that undergoes the other function.

For example: if f(x) is the inverse of g(x) then;

for every x.

As per the statement:

If h(x) is the inverse of f(x).

by definition of inverse:

Therefore, the value of h(f(x)) is, x

6 0

4 years ago

Read 2 more answers

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