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Anna [14]
2 years ago
5

It's about "rates and unit rates" sorry again I also need help here sorry I'm learning something new and I need a examples

Mathematics
1 answer:
san4es73 [151]2 years ago
5 0

Answer:

1. D

2. A

I think that is the answer

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John has 367 stamps And his uncle gave him some more stamps now he have 462 stamps how many stamps did he get from his uncle
Ulleksa [173]

Answer:

The answer is 829 because you are adding

Step-by-step explanation:

7 0
3 years ago
-3(y-5) simplify please
d1i1m1o1n [39]
Im pretty sure its 8y because two negatives equal a positive so 5 plus 3 and then you isolated the y so 8y. :)
5 0
2 years ago
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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.

6 0
3 years ago
Determine whether the polynomial is a difference of squares and if it is, factor it.
ladessa [460]

Answer:

(y - 4)(y + 4)

Step-by-step explanation:

y² - 16 ← is a difference of squares

Since y² and 16 are both squares separated by a difference , that is minus

A difference of squares factors as

y² - 16

= y² - 4²

= (y - 4)(y + 4)

5 0
3 years ago
Find the value of x for which the lines a and b are parallel.
siniylev [52]

Answer:

x = 19

Step-by-step explanation:

148 and 7x+15 are alternate interior angles and alternate interior angles are equal when the lines are parallel

148 = 7x+15

Subtract 15 from each side

148-15 = 7x+15-15

133 =7x

Divide each side by 7

133/7 =7x/7

19=x

4 0
3 years ago
Read 2 more answers
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