A tree in Chau's backyard has been growing for 2.4 years. The tree is .25 meters tall.

A pattern follows the rule "Starting with two, every consecutive line has a number one more than the previous line."

mars1129 [50]

There will be eight marbles on the 9th line

3 0

3 years ago

Reason Abstractly Find two integers whose sum is -7 and whose

Travka [436]

Step-by-step explanation:

Let's represent the two integers with the variables $x$ and $y$ .

From the problem statement, we can create the following two equations:

$x + y = -7$

$xy = 12$

With the first equation, we can subtract $y$ from both sides to isolate the $x$ variable to the left-hand side:

$x = -7 - y$

Now that we have a value for $x$ , we can plug it into the second equation and solve for $y$ :

$(-7 - y)y = 12$

$-7y - y^{2} = 12$

Now, let's move everything to one side of the equation:

$y^{2} + 7y + 12 = 0$

Factoring this quadratic will give us two values for $y$ :

$(y + 4)(y + 3) = 0$

$y = -3, -4$

Since we now know $y = -3, -4$ , we can plug this back into either of the original equations to get a value for $x$ , which will be $x = -4, -3$ .

So the two numbers that sum to $-7$ and have a product of $12$ are $-3, -4$ .

7 0

3 years ago

Read 2 more answers

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

If the volume of a box is 2x3 + 4x2 − 30xwhich of the dimensions are possible with the given x-value?

Kipish [7]

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$ .

Step-by-step explanation:

The given is,

$2 x^{3} + 4 x^{2} -30x$ ................................(1)

Step:1

Check for option A,

x = 1, dimensions 8 by 9 by 1

From the equation (1),

Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

$=2+4-30$ = -24...................(2)

From the dimensions,

Volume = ( 8 × 9 × 1 )

= 72............................................(3)

From equation (2) and (3)

-24 ≠ 72

So, X=1; dimensions 8 by 9 by 1 is not possible.

Check for option B,

x = 1, dimensions 2 by 5 by 3

From the equation (1),

Volume = $2 (1^{3}) + 4 (1^{2} )-30(1)$

$=2+4-30$ = -24...................(4)

From the dimensions,

Volume = ( 2 × 5 × 3 )

= 30.........................................(5)

From equation (4) and (5)

-24 ≠ 30

So, X=1; dimensions 2 by 5 by 3 is not possible.

Check for option C,

x = 4, dimensions 2 by 5 by 3

From the equation (1),

Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

$=2(64)+4(16)-30(4)$

$= 128+64-120$

= 72.............................................(6)

From the dimensions,

Volume = ( 2 × 5 × 3 )

= 30............................................(7)

From equation (6) and (7)

72 ≠ 30

So, X=4; dimensions 2 by 5 by 3 is not possible.

Check for option C,

x = 4, dimensions 8 by 9 by 1

From the equation (1),

Volume = $2 (4^{3}) + 4 (4^{2} )-30(4)$

$=2(64)+4(16)-30(4)$

$= 128+64-120$

= 72............................................(8)

From the dimensions,

Volume = ( 8 × 9 × 1 )

= 72............................................(9)

From equation (8) and (9)

72 = 72

So, X=4; dimensions 8 by 9 by 3 is possible.

Result:

The possible value of x = 4, dimensions 8 by 9 by 1 (option D), if the volume of a box is $2 x^{3} + 4 x^{2} -30x$ .

4 0

3 years ago

Use Graphs A-D for Exercises 9-11. During the winter, the amount of water that flows

Morgarella [4.7K]

did u get the awnsers?

6 0

3 years ago