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3 years ago

The aspect ratio is used when calculating the aerodynamic efficiency of the wing of a plane. For a standard wing area, the

Mathematics

2 answers:

tatyana61 [14]3 years ago

8 0

Answer: y=s^2/36 and y=5.7;s=14.3ft or A

Step-by-step explanation:

xxMikexx [17]3 years ago

7 0

Answer:

Y=s^2/36 and y=5.7;14.3 ft

Step-by-step explanation:

The question was not typed correctly. Here, a better version:

The aspect ratio is used when calculating the aerodynamic efficiency of the wing of a plane for a standard wing area, the function A(s)=s^2/36 can be used to find the aspect ratio depending on the wingspan in feet. If one glider has an aspect ratio of 5.7, which system of equations and solution can be used to represent the wingspan of the glider? Round solution to the nearest tenth if necessary.

Y=s^2/36 and y=5.7;14.3 ft

Y=5.7s^2 and y=36; s=2.5ft

Y=36s^2 and y=0; s=0.4 ft

Y=s^2/36 +5.7 and y=0; s=5.5 ft

In the function A(s)=s^2/36 A(s) represents the aspect ratio and s the wingspan. If one glider has an aspect ratio of 5.7, then A(s) = 5.7. We want to know the wingspan of the glider. Replacing A(s) by Y we get the following system of equation:

Y=s^2/36

with y = 5.7

5.7 = s^2/36

5.7*36 = s^2

√205.2 = s

14.3 ft

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$444-66%= $150.96= $151

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2 years ago

Which inequality comparing two of the expressions below is true when 0.5

Aleonysh [2.5K]

Answer:

Option C

Step-by-step explanation:

Given 0.5≤ x ≤ 0.7

According to the range of x, we will check which option is true

Let x = 0.6

A. $\frac{x}{3} > \sqrt{x}$

If x = 0.6

∴0.6/3 > √0.6 ⇒ 0.2 > 0.77 ⇒ Wrong inequality

B. x³ > 1/x

If x = 0.6

∴ 0.6³ > 1/0.6 ⇒ 0.216 > 1.667 ⇒ Wrong inequality

C. √x > x³

If x = 0.6

∴ √0.6 > 0.6³ ⇒ 0.77 > 0.216 ⇒ True inequality

D. (x/3) > (1/x)

If x = 0.6

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So, The true inequality is option C

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2 years ago

Simplify

denis23 [38]

Hello!

Answer:

1. =-6

2. =39

3. =-603

_________________________________________________

Explanation: This question's it's about the order of operations. It stands for p-parenthesis, e-exponents, m-multiply, d-divide, a-add, and s-subtract. It can also go left to right. (PEMDAS)

1. -4(3-1)+2

Do parenthesis first.

$(3-1)=2$

$-4*2+2$

Then do multiply left to right.

$-4*2=-8$

Add left to right.

$-8+2$

$8-2=6$

*Answer should be have negative sign.*

Answer: → $\boxed{=-6}$

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2. 3⋅[(30−8)÷2+2]

Do parenthesis first.

$(30-8)=22$

$22/2+2$

Then divide left to right.

$22/2=11$

Add left to right.

$11+2=13$

Multiply left to right.

$3*13=39$

*Answer should be have positive sign.*

Answer: → $\boxed{=39}$

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3. 14(1-23)2+13

Do parenthesis first.

$(1-23)=-22$

$14(-22)*2+13$

Then, you multiply left to right.

$14(-22)=-308$

$-308*2=-616$

Add/subtract left to right.

$-616+13$

$616-13=603$

*The answer must be have a negative sign*

Answer: → $\boxed{=-603}$

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Hope this helps!

Thank you!

Have a great day!

-Charlie

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2 years ago

For the polynomial function ƒ(x) = −x6 + 3x4 + 4x2, find the zeros. Then determine the multiplicity at each zero and state wheth

murzikaleks [220]

There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.

You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.

-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)

Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.

-x^2(x^4 - 3x^2 - 4)

-x^2(x^2 - 4)(x^2 + 1)

Now we can use the factors of two perfect squares rule to factor the middle parenthesis.

-x^2(x^2 - 4)(x^2 + 1)

-x^2(x - 2)(x + 2)(x^2 + 1)

We would also want to split the term in the front.

-x^2(x - 2)(x + 2)(x^2 + 1)

(x)(-x)(x - 2)(x + 2)(x^2 + 1)

Now we would set each portion equal to 0 and solve.

First root

x = 0 ---> no work needed

Second root

-x = 0 ---> divide by -1

x = 0

Third root

x - 2 = 0

x = 2

Forth root

x + 2 = 0

x = -2

Fifth and Sixth roots

x^2 + 1 = 0

x^2 = -1

x = +/- $\sqrt{-1}$

x = +/- i

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3 years ago

The area of the square

maria [59]

Answer:

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Step-by-step explanation:

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3 years ago

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