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3 years ago

What is the slope of the tangent line through (0,-3) and (x,y) for y = 3x2 - 3 and x=1?

1 answer:

4 0

First you must derive the equation to find the slope:
y = 3x ^ 2 - 3
y '= 6x
Then, for x = 1 we have the slope is:
y '(1) = 6 (1) = 6
Then the equation of the line with slope equal to 6 and passing through point (0, -3) is:
y = 6x - 3
The slope is m = 6

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Set up and solve a system of equations to solve the problem.

NeTakaya

Answer:

The boys club sold 52 bags of hardwood mulch and 84 bags of pine bark mulch

Step-by-step explanation:

Let

x ----> the number of bags of hardwood mulch sold

y ----> the number of bags of pine bark mulch sold

we know that

A local boys club sold 136 bags of mulch and made a total of $400

$x+y=136$ ----> equation A

$3.25x+2.75y=400$ ----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

Using a graphing tool

The solution is the point (52,84)

see the attached figure

therefore

The boys club sold 52 bags of hardwood mulch and 84 bags of pine bark mulch

6 0

3 years ago

Find the distance between the points (4, 3) and (0, 3).

Serga [27]

The answer is 5 points

4 0

2 years ago

Read 2 more answers

What is two over twelve + four over nine

noname [10]

Answer:

2/3

Step-by-step explanation:

2/12+4/9

Find a common denominator

16/72+ 32/72

48/72

Simplify

=2/3

Hope this helps

3 0

3 years ago

80% of what(?) games is 32 games..?

exis [7]

Answer=40 games

32games x games
________=________
80% 100%

Cross Multiply
80x=3200
divide both sides by 80
x=40games

80% of 40games is 32 games

3 0

3 years ago

Read 2 more answers

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

5 0

2 years ago