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Lynna [10]
2 years ago
11

Which of the following equations demonstrate the multiplication property of equality for the equation x = 2? Check all

Mathematics
1 answer:
igor_vitrenko [27]2 years ago
7 0

Answer: x(5) = 10 has the multiplication property.

Step-by-step explanation:

To verify the multiplication property of equality , We need to put  the value of x= 2

1.  x(5) = 10

Solution = 2(5) = 10     ✔

2.  x(1.5) = 2(2)

Solution = 2 (1.5) = 3    X

3.  x(0.5) = 2(0.5)

Solution = 2 (0.5) = 1    X

4.  x(12) = 12

Solution = 2 (12) = 24    X

For more multiplication property related question visit here

brainly.com/question/14059007

#SPJ1

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Find the slope of the line that passes through the points (1,-4) and (3,-1).
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Answer: E

Step-by-step explanation: I used m.athway

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3 years ago
Find the interval in which f(x)=sinx−cosx is increasing or decreasing?
alisha [4.7K]

Answer:

There is no short answer.

Step-by-step explanation:

To find to intervals which f(x) increases or decreases, we first need to find it's derivative.

f(x) = sinx - cosx\\f'(x) = cosx - (-sinx) = cosx + sinx

The function is increasing when it's value is  > 0 and decreasing when it's value is < 0.

If we take a look at this graph, cosx+sinx is positive when they are both positive or when cosx is greater then sinx on the negative part.

I hope this answer helps.

3 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.

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(m-21) (m+1)there I thinks it's right unless I messed up my work

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