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AnnyKZ [126]
3 years ago
9

Please help I have been stuck on these two for two days now!!!

Mathematics
1 answer:
postnew [5]3 years ago
8 0

First of all you should demand your teachers stop teaching you algebra in geometry class.  

1

Linear pairs (and this is the geometry part) are angles that add to 180 degrees, supplementary angles formed by two lines crossing.

3n + 21 + 2n + 34 = 180

5n + 55 = 180

5n = 125

n = 25

mEFG = 3n + 21 = 96 degrees

mCFH = 2n + 34 = 84 degrees

These are indeed supplements, adding to 180 degrees.

2

ACD is a right angle because it makes a linear pair along DF with another right angle.

So

mACB + mBCD = 90 degrees

mBCD = 90 - mACB = 90 - 68

mBCD = 22 degrees

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3 years ago
9/6 × 4/5 <br> 1 <br>) 36/11<br>) 6/5<br>) 32/30<br>) 13/15​
erik [133]

Answer:

6/5

Step-by-step explanation:

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6 0
2 years ago
Solve the following quadratic equations by any method. The solutions are:
Nataly_w [17]

Answer:

Option C .

Step-by-step explanation:

We would like to solve the below <u>quadratic </u><u>equation</u><u> </u>,

\longrightarrow f(x) = x^2-12x +32

Step 1 : <u>E</u><u>q</u><u>u</u><u>a</u><u>t</u><u>e</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u><u>w</u><u>i</u><u>t</u><u>h</u><u> </u><u>0</u><u> </u><u>:</u><u>-</u>

\longrightarrow f(x) = 0\\

\longrightarrow x^2-12x + 32=0

Step 2 : <u>F</u><u>a</u><u>c</u><u>t</u><u>o</u><u>r</u><u>i</u><u>s</u><u>e</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>R</u><u>H</u><u>S</u><u> </u><u>:</u><u>-</u>

\longrightarrow x^2-8x - 4x +32=0\\

\longrightarrow x(x-8)-4(x-8)=0\\

\longrightarrow (x-8)(x-4)=0\\

Step 3 : <u>E</u><u>q</u><u>u</u><u>a</u><u>t</u><u>e</u><u> </u><u>e</u><u>a</u><u>c</u><u>h</u><u> </u><u>f</u><u>a</u><u>c</u><u>t</u><u>o</u><u>r</u><u> </u><u>w</u><u>i</u><u>t</u><u>h</u><u> </u><u>0</u><u> </u><u>:</u><u>-</u>

\longrightarrow x - 8 = 0\\ \qquad x -4=0

\longrightarrow x = 8 \qquad x = 4

\longrightarrow \underline{\underline{\boldsymbol{ x = 8,4}}}

<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u> </u><u>o</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>C</u><u> </u><u>i</u><u>s</u><u> </u><u>c</u><u>o</u><u>r</u><u>r</u><u>e</u><u>c</u><u>t</u><u> </u><u>.</u>

8 0
2 years ago
Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2
Anastaziya [24]

Answer:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

Step-by-step explanation:

we would like to figure out the derivative of the following:

\displaystyle  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

to do so, let,

\displaystyle y =  \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }

By simplifying we acquire:

\displaystyle y =  3 -  \frac{2}{x}  -  \frac{1}{ {x}^{2} }

use law of exponent which yields:

\displaystyle y =  3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} }

take derivative in both sides:

\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  (3 -  2 {x}^{ - 1}  -   { {x}^{  - 2} } )

use sum derivation rule which yields:

\rm\displaystyle  \frac{dy}{dx} =  \frac{d}{dx}  3 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

By constant derivation we acquire:

\rm\displaystyle  \frac{dy}{dx} =  0 -   \frac{d}{dx} 2 {x}^{ - 1}  -     \frac{d}{dx} {x}^{  - 2}

use exponent rule of derivation which yields:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ - 1 -1} ) -     ( - 2 {x}^{  - 2 - 1} )

simplify exponent:

\rm\displaystyle  \frac{dy}{dx} =  0 -   ( - 2 {x}^{ -2} ) -     ( - 2 {x}^{  - 3} )

two negatives make positive so,

\displaystyle  \frac{dy}{dx} =   2 {x}^{ -2} +      2 {x}^{  - 3}

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

\displaystyle  \frac{dy}{dx} =   \frac{2 }{x^2}+       \frac{2}{x^3}

simplify addition:

\displaystyle  \frac{dy}{dx} =    \frac{2x + 2}{x^3}

and we are done!

5 0
3 years ago
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