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3 years ago

15

9

x} ) ^5 " align="absmiddle" class="latex-formula"> How do you rewrite the following radical expression in rational exponent form?

2 answers:

BlackZzzverrR [31]3 years ago

8 0

<span>How do you rewrite the following radical expression in rational exponent form?
</span>9 1/2 is the answer

Maksim231197 [3]3 years ago

3 0

9 1/2 lol you're welcome I can see you and I hate him

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A total of $12,000 is invested at two different rates: $5,000 of it is invested at 6%, and the rest is invested at 8%. How much

lesantik [10]

5000*6/100=300
7000*8/100=560
total=860

7 0

3 years ago

Pls!!!!!!!!!!!!!!!!!!!!

lesya692 [45]

Answer:

C

Step-by-step explanation:

The table shows a linear function.

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (1, 3) and (x₂, y₂ ) = (2, 7) ← 2 ordered pairs from the table

m = $\frac{7-3}{2-1}$ = 4, thus

y = 4x + c ← is the partial equation

To find c substitute any ordered pair into the partial equation

Using (1, 3), then

3 = 4 + c ⇒ c = 3 - 4 = - 1

y = 4x - 1 → C

3 0

3 years ago

Need help please.........thank you

crimeas [40]

A²-b²=(a-b)(a+b)
4x² - 25= (2x)²- 5² =(2x-5)(2x+5)
You do not have this choice, so the answer is D.None of the above.

6 0

3 years ago

Read 2 more answers

an isosceles triangle has congruent sides of 20cm. the base is 10cm. find the height of the triangle.

dolphi86 [110]

Answer:

$\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}$

Step-by-step explanation:

Look at the picture.

The formula of an area of a triangle:

$A_\triangle=\dfrac{bh}{2}$

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

<em />

We need a length of a height.

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have:

$leg=5,\ leg=h,\ hypotenuse=20$

Substitute:

$5^2+h^2=20^2$

$25+h^2=400$ <em>subtract 25 from both sides</em>

$h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm$

Calculate the area:

$A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2$

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