<img src="https://tex.z-dn.net/?f=x-7%20%3D%2010" id="TexFormula1" title="x-7 = 10" alt="x-7 = 10" align="absmiddle" class="late

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Marina CMI [18]

3 years ago

x-formula">
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Mathematics

2 answers:

Artist 52 [7]3 years ago

3 0

Answer:

This is your answer <(￣︶￣)↗

Step-by-step explanation:

plz mark me a brainlist..

#hope it helps you..

(◕ᴗ◕)

TEA [102]3 years ago

3 0

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

$x - 7 = 10 \\ \\ x = 10 - 7 \\ \\ x = 3$

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Solve 2x^3+7x^2-2x-1=o

Mashcka [7]

Answer:

$x=\sqrt{3}-2\\x=-\sqrt{3}-2\\x=\frac{1}{2}$

Step-by-step explanation:

This is a hard one

We have to use the rational root theorem

$2x^3+7x^2-2x-1$ = 0

We have to find all the factors of a and d and put them in a fraction

$a=2\\d=-1\\\frac{1}{1,2}$

We then plug them into the equation to see if any of them work

The equation isn't true when plugging 1, but is true when plugging in 1/2

factored form of 1/2 is (2x-1)

Then we divide the original equation by (2x-1) (you can use synthetic division or long division, it would be hard to type out the process for that) to get $x^2+4x+1$

So now the equation is $(2x-1)(x^2+4x+1)$

Solve the second half of this equation using the quadratic formula to get

$\sqrt{3}-2$ and $-\sqrt{3}-2$

We already know the solution for the first half of the equation (1/2)

So the final answers are:

$x=\sqrt{3}-2, -\sqrt{3}-2, \frac{1}{2}$

8 0

3 years ago

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft and on the other three s

makvit [3.9K]

Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = $Length \times Breadth = xy$

We are given that the area of the garden is 122 square feet

So, $xy=122$ ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

Substitute the value of y from A

Total cost = $30x+20(\frac{122}{x})$

Total cost = $\frac{2440}{x}+30x$

Now take the derivative to minimize the cost

$f(x)=\frac{2440}{x}+30x$

$f'(x)=-\frac{2440}{x^2}+30$

Equate it equal to 0

$0=-\frac{2440}{x^2}+30$

$\frac{2440}{x^2}=30$

$\sqrt{\frac{2440}{30}}=x$

$9.018 =x$

Now check whether it is minimum or not

take second derivative

$f'(x)=-\frac{2440}{x^2}+30$

$f''(x)=-(-2)\frac{2440}{x^3}$

Substitute the value of x

$f''(x)=-(-2)\frac{2440}{(9.018)^3}$

$f''(x)=6.6540$

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

$(9.018)y=122$

$y=\frac{122}{9.018}$

$y=13.528$

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

4 0

3 years ago

- 7th Grade Work -<br><br> I rlly need help ; - ;

Tanya [424]

Answer:

The length would be 1.5x + 5.

Step-by-step explanation:

So first with this, note that 2 length + 2 height gives the perimeter of the rectangle.

Given this:

2(1/2x + 4) + 2(?) = 4x+18

x+8 + 2(?) = 4x+18

2(?) = 3x+10

? = 1.5x+5

1.5x + 5 = 1.5x + 5

The length would be 1.5x + 5.

To double check, by plugging in 1.5x + 5 back into the equation for the perimeter, you will get the same perimeter:

2(1.5x + 5) + 2(1/2x+4) = P

3x+10 + x+8 = P

4x + 18 = P

8 0

3 years ago

Find the term to term rules for the following sequences. 28 25 22 19 16 .

Mekhanik [1.2K]

Answer:

a) next one is previous one -3

b) next one is 3 times previous one

3 0

3 years ago

Read 2 more answers

Which of the following statements is true for the logistic differential equation?

solong [7]

Answer:

All of the above

Step-by-step explanation:

dy/dt = y/3 (18 − y)

0 = y/3 (18 − y)

y = 0 or 18

d²y/dt² = y/3 (-dy/dt) + (1/3 dy/dt) (18 − y)

d²y/dt² = dy/dt (-y/3 + 6 − y/3)

d²y/dt² = dy/dt (6 − 2y/3)

d²y/dt² = y/3 (18 − y) (6 − 2y/3)

0 = y/3 (18 − y) (6 − 2y/3)

y = 0, 9, 18

y" = 0 at y = 9 and changes signs from + to -, so y' is a maximum at y = 9.

y' and y" = 0 at y = 0 and y = 18, so those are both asymptotes / limiting values.

8 0

3 years ago