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Rudiy27
3 years ago
14

Solve the equation 8x-12=5+15

Mathematics
2 answers:
Anika [276]3 years ago
7 0
8x-12 = 5+15
8x-12 = 20 
     8x = 32      +12 
       x = 4        ÷8

maxonik [38]3 years ago
3 0
8x-12=5+15       Regroup
    +12  +12

8x= 5+15+12       Add like terms

8x=5+15+12

8x=32               Divide both sides by 8
/8    /8

 x=4 is your answer



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Mia's soccer coach told her to keep track of how much water she drinks. Yesterday she drank 18 ounces with breakfast, 21 ounces
igor_vitrenko [27]
Yes, 82 ounces is the correct answer.
6 0
3 years ago
g Use this to find the equation of the tangent line to the parabola y = 2 x 2 − 7 x + 6 at the point ( 4 , 10 ) . The equation o
natali 33 [55]

Answer:

The tangent line to the given curve at the given point is y=9x-26.

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of y=2x^2-7x+6 and then evaluate it for x=4.

(y=2x^2-7x+6)'          Differentiate the equation.

(y)'=(2x^2-7x+6)'       Differentiate both sides.

y'=(2x^2)'-(7x)'+(6)'    Sum/Difference rule applied: (f(x)\pmg(x))'=f'(x)\pm g'(x)

y'=2(x^2)'-7(x)'+(6)'  Constant multiple rule applied: (cf)'=c(f)'

y'2(2x)-7(1)+(6)'        Applied power rule: (x^n)'=nx^{n-1}

y'=4x-7+0               Simplifying and apply constant rule: (c)'=0

y'=4x-7                    Simplify.

Evaluate y' for x=4:

y'=4(4)-7

y'=16-7

y'=9 is the slope of the tangent line.

Point slope form of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on the line.

Insert 9 for m and (4,10) for (x_1,y_1):

y-10=9(x-4)

The intended form is y=mx+b which means we are going need to distribute and solve for y.

Distribute:

y-10=9x-36

Add 10 on both sides:

y=9x-26

The tangent line to the given curve at the given point is y=9x-26.

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function f(x)=2x^2-7x+6 at x=4 (the slope of the tangent line at x=4):

\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}

\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}  We are given f(4)=10.

\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

2x^2-7x-4=(x-4)(2x+1)

Let's check this with FOIL:

First: x(2x)=2x^2

Outer: x(1)=x

Inner: (-4)(2x)=-8x

Last: -4(1)=-4

---------------------------------Add!

2x^2-7x-4

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}

\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}

\lim_{x \rightarrow 4}(2x+1)

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

6 0
3 years ago
1/2 - 3 1/4<br> 1/2 - 4/5<br> 1/6 + -3 3/4
Olenka [21]

Answer:

1/2 - 3 1\4= -11/4

1\2 - 4\5= -3\4

1\6 + -3 3\4= 43\12

3 0
3 years ago
In the figure above, AB is tangent to the circle with center O
Radda [10]

<u>Given</u>:

Given that O is the center of the circle.

AB is tangent to the circle.

The measure of ∠AOB is 68° and we know that the tangent meets the circle at 90°

We need to determine the measure of ∠ABO.

<u>Measure of ∠ABO:</u>

The measure of ∠ABO can be determined using the triangle sum property.

Applying the property, we have;

\angle ABO+\angle BAO+\angle AOB=180^{\circ}

Substituting the values, we get;

\angle ABO+90^{\circ}+68^{\circ}=180^{\circ}

Adding the values, we have;

\angle ABO+158^{\circ}=180^{\circ}

Subtracting both sides by 158, we get;

\angle ABO=22^{\circ}

Thus, the measure of ∠ABO is 22°

6 0
3 years ago
Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]

3 0
3 years ago
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