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2 years ago

Plz help !!!!!!!!!!!

Mathematics

2 answers:

NNADVOKAT [17]2 years ago

4 0

Answer:

the first one

Step-by-step explanation:

the 36 and the 5 stay in the same places

the x with the negative 4 exponent moves to the bottom and combines with x squared so that makes x to the 6th power on the bottom.

The y and x move to the top to combine with what is there.

vodomira [7]2 years ago

3 0

Answer: $\bold{a)\quad \dfrac{36y^5z^2}{5x^6}}$

Step-by-step explanation:

$\dfrac{36x^{-4}y^2z^0}{5x^2y^{-3}z^{-2}}\\\\\\=\dfrac{36x^{-4+4}y^{2+3}z^{0+2}}{5x^{2+4}y^{-3+3}z^{-2+2}}\\\\\\=\dfrac{36x^{0}y^5z^2}{5x^6y^{0}z^{0}}\\\\\\=\dfrac{36y^5z^2}{5x^6}$

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Can someone help me with this question ? Show steps please

jek_recluse [69]

Answer:

B) $\boxed{A_T=108\;in^2}$

Step-by-step explanation:

That is a triangular prism and thus, it has two bases that are triangles, in this case, right triangles, whose hypotenuse is 5 in. long and one of its legs is 4 in. and the other one, 3 in.

The first step is to find the area of both bases like this:

$A_B=2\left(\dfrac{b\cdot h}{2}\right)=b\cdot h=4\cdot 3\\\\ \boxed{A_B=12\;in^2}$

The second step is to get the lateral area, which is compound of three rectangles, hence the process is the following:

$A_L=b_1\cdot h_1+b_2\cdot h_2+b_3\cdot h_3\\A_L=8\cdot3+8\cdot4+8\cdot5\\A_L=24+32+40\\\boxed{A_L=96\;in^2}$

Finally, we sum both of the areas found to calculate the total area.

$A_T=A_B+A_L\\A_T=12+96\\\boxed{A_T=108\;in^2}$

So, the answer is option B).

4 0

2 years ago

| If f(x) = (3x – 1, find f(2) and f(-2)

Brums [2.3K]

Answer:

f(2)=5, f(-2)=-7

Step-by-step explanation:

Step 1 : f(2)=3(x)-1

f(2)=5

Step 2 : f(-2)=3(x)-1

f(-2)=-7

6 0

2 years ago

HELP ASAP Simplify (xy3z,4)4 A. xy3z16 B. xy3z8 C. x4y12z16 D. x5y7z8

MariettaO [177]

Answer:

(xy^3z,4)

Step-by-step explanation:

3 0

2 years ago

Help me please, thank you! :)

EastWind [94]

Answer:

It's 16 units. Just look at the lengths and how many times they are on a shape. The shape is a rectangle, which means there will be two of each length, so just add them all together. 6+6+2+2=16

6 0

2 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

2 years ago