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2 years ago

Brainly will be given out.

Mathematics

2 answers:

Oksana_A [137]2 years ago

6 0

Step-by-step explanation:

I don't get what your asking.

igomit [66]2 years ago

6 0

I’m confused on what ur asking

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The shape is a Rectangle

Gnoma [55]

8 times 17 is 136
so 8 is your answer

3 0

3 years ago

Solve the following quadratic equation by using the quadratic formula.

Bas_tet [7]

Answer:

B. $\frac{-3}{4}$ ± $\frac{\sqrt{41} }{4}$

Step-by-step explanation:

First, substitute the equation into the quadratic formula.

$x=\frac{b+-\sqrt{(b^{2} )-4(a)(c)} }{2a}$

x= $\frac{-3+-\sqrt{(3)^2-4(-2)(4)} }{4}$

Solve for x.

x= $\frac{-3+-\sqrt{9+32} }{4}$

x= $\frac{-3+-\sqrt{41} }{4}$

The answer is x= $-\frac{3}{4}$ ± $\frac{\sqrt{41} }{4}$

6 0

3 years ago

Use the quadratic formula to find both solutions to the quadratic equation given below 3x^2-x+6=0

Mrac [35]

Hello from MrBillDoesMath!

Answer:

Choice E and F

Discussion:

From the quadratic formula with a = 3, b = -1, and c = 6

x = ( -b +\- sqrt(b^2 - 4ac) ) / (2a) => substitute in a,b,c from above

x = ( -(-1) +\- sqrt((-1)^2 - 4(3)(6)) / (2*3) => discriminant = 1 - 72 = -71)

x = ( 1 +\- sqrt(-71))/ 6

which are choices E and F

Thank you,

MrB

6 0

4 years ago

Read 2 more answers

A piece of wire of length 5050 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the

algol [13]

Answer:

x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

Step-by-step explanation:

Let x be the length of wire that is cut to form a circle within the 5050 wire, so 5050 - x would be the length to form a square.

A circle with perimeter of x would have a radius of x/(2π), and its area would be

$A_c = \pir^2 = \pi (\frac{x}{2pi})^2 = \frac{x^2}{4\pi}$

A square with perimeter of 5050 - x would have side length of (5050 - x)/4, and its area would be

$A_s = (\frac{5050 - x}{4})^2 = \frac{(5050 - x)^2}{16}$

The total combined area of the square and circles is

$\sum A = A_c + A_s = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$

To find the maximum and minimum of this, we just take the 1st derivative, and set it to 0

$A' = \frac{2x}{4\pi} + \frac{-2(5050-x)}{16} = 0$

$\frac{x}{2\pi} - \frac{5050 - x}{8} = 0$

Multiple both sides by 8π and we have

$4x - 5050\pi + x\pi = 0$

$x(4 + \pi) = 5050\pi$

$x = \frac{5050\pi}{4 + \pi} = 2221.5$

At x = 2221.5:

$A = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$ = 392720 + 500026 = 892746 [/tex]

At x = 0, $A = 5050^2/16 = 1593906$

At x = 5050, $A = \frac{5050^2}{4\pi} = 2029424$

As 892746 < 1593906 < 2029424, x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

3 0

3 years ago

Wha is the value of x(3) + y(3) when<br> 3 and y = 4?

nordsb [41]

Answer:

Step-by-step explanation:

3 x 3 = 9

4 x 4 = 16

9 + 16 = 25

6 0

3 years ago