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3 years ago

I need to put equations into slope intercept and i need big help its a test grade

Mathematics

1 answer:

creativ13 [48]3 years ago

5 0

Answer:

y= 2/3x + 3

Step-by-step explanation:

By using the distributive property on the right side of the equation you get 2/3x + 2. Then you add 1 to the right side to cancel out the (-1) on the left.

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Using differential calculus, maximize the volume of a box made of cardboard (top is open) as shown in Figure A. 15, subject to t

lana [24]

The maximum volume of the box is 40√(10/27) cu in.

Here we see that volume is to be maximized

The surface area of the box is 40 sq in

Since the top lid is open, the surface area will be

lb + 2lh + 2bh = 40

Now, the length is equal to the breadth.

Let them be x in

Hence,

x² + 2xh + 2xh = 40

or, 4xh = 40 - x²

or, h = 10/x - x/4

Let f(x) = volume of the box

= lbh

Hence,

f(x) = x²(10/x - x/4)

= 10x - x³/4

differentiating with respect to x and equating it to 0 gives us

f'(x) = 10 - 3x²/4 = 0

or, 3x²/4 = 10

or, x² = 40/3

Hence x will be equal to 2√(10/3)

Now to check whether this value of x will give us the max volume, we will find

f"(2√(10/3))

f"(x) = -3x/2

hence,

f"(2√(10/3)) = -3√(10/3)

Since the above value is negative, volume is maximum for x = 2√(10/3)

Hence volume

= 10 X 2√(10/3) - [2√(10/3)]³/4

= 2√(10/3) [10 - 10/3]

= 2√(10/3) X 20/3

= 40√(10/27) cu in

To learn more about Maximization visit

brainly.com/question/14682292

#SPJ4

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3 0

1 year ago

Find the distance between the origin and the points (-8,4)

Papessa [141]

Hi, I'm happy to help!

To solve this, we need to find the distance from the origin to the y coordinate value, x coordinate value, then use the Pythagorean Theorem.

The origin of a graph (center) has the coordinates (0,0), so this will be our other coordinates.

First, let's find the x coordinate distance change. We move from the x coordinate 0, to the x coordinate -8, so we move 8 x units.

Next, let's find the y coordinate distance change. We move from the y coordinate 0, to the y coordinate 4, so we move 4 y units.

Now that we have these two leg lengths, let's imagine this as a right triangle. Moving from the origin, we draw a line from (0,0), to (-8,0). Then, we draw a line from (-8,0) to (-8,4). Now, draw a direct line from (0,0), to (-8,4). We have the length of the first(8) and second(4) lines, and we need to find the third line length to find our answer. To do this, we use the Pythagorean Theorem, which states that a²+b²=c². This says that, in a right triangle, the square of the two shorter lengths equals the square of the longest length. The longest length is what we are solving for.

Let's say x distance is a, and y distance is b. Now, apply the values:

8²+4²=c²

64+16=c²

80=c²

Now, we need to find the value of c, so we need to find the square root of 80.

√80=c

8.9442...=c

Since the number goes on forever, we need to round it. For this example, let's round it to the nearest tenth, which would be 8.9.

To summarize the distance between the origin and the coordinates (-8,4), is about 8.9.

I hope this was helpful, keep learning! :D

8 0

2 years ago

−3.5=x4 please help me

Studentka2010 [4]

The answer is -0.875

3 0

3 years ago

What is the distance between point A and point B? Enter your answer in the box.

finlep [7]

The answer to this question is= 13

4 0

3 years ago

Read 2 more answers

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago