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daser333 [38]
3 years ago
8

What is six and five hundredths in standard form? 6.5 6.005 6,500 6.05

Mathematics
1 answer:
dsp733 years ago
8 0
6.05 because the 5 is in the hundredths place...
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Tasha claims that her skateboard weighs about 3 pounds. is Tasha's claim reasonable? Explain
gladu [14]
Yes, It would be reasonable since thats about the weight of a skateboard.
However, the average skateboard is closer to 6 pounds. So, the skateboard in the question is very possible- but it would be really low quality.

However- given that you are in middle school- I would say yes, it is reasonable.
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2 years ago
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Is this equation a linear relation, and please solve the equation?<br> -3x + 2y = 1
Liono4ka [1.6K]

Answer:

Step-by-step explanation:

It is a linear equation and the gradient equation is given as:

y=mx+c

Where m is the gradient and c is the y intercept

-3x+2y=1

2y=1+3x

Dividing all through by 2

y=1/2 + 3x/2

Therefore the gradient of this linear equation is 3/2 and the y intercept is 1/2

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3 years ago
Find the distance represented of the coordinate plane
antiseptic1488 [7]

Answer:

The answer to your question is: letter A

Step-by-step explanation:

From the graph we get the points,

P (2,1)

Q (6,8)

Formula

d = √((x2-x1)² + (y2-y1)²)

d = √((6-2)² + (8-1)²)

d = √ (4² + 7²)

d = √ (16 + 49

d = √65           letter A

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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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Answer : C

6 2 over 3 units
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