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3 years ago

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 1/2 - 4/5 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]

Given:

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.

Measure of ∠ABO:

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