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2 years ago

Please help me with this question! i will give brainiest if correct! (if you show me how to lol i have no idea how)

Mathematics

1 answer:

Nitella [24]2 years ago

7 0

Answer:

y(2x² + y) / x(5y² - 6x)

Step-by-step explanation:

Simplify the numerator:

1/x² + 2/y =
y/(x²y) + 2x²/(x²y) =
(2x² + y)/(x²y)

Simplify the denominator:

5/x - 6/y² =
5y²/(xy²) - 6x/(xy²) =
(5y² - 6x) / (xy²)

Simplify the fraction:

(2x² + y)/(x²y) ÷ (5y² - 6x) / (xy²) =
(2x² + y)/(x²y) × xy² / (5y² - 6x) =
y(2x² + y) / x(5y² - 6x)

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2 · 32 + (3 · 2)2. i need the answer

vfiekz [6]

Answer: 76

Step-by-step explanation:

3 * 2 = 6

6* 2 = 12

2 * 32 = 64

12 + 64 = 76

4 0

3 years ago

Can you solve this integral?<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5Ea_b%20%7B%20x%5E%7B2%7D%20%7D%20%5C%2C%20dx%2

adelina 88 [10]

$\frac{1}{3}(a^3-b^3)$

4 0

3 years ago

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two bakeries' specialty pi

DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0

3 years ago

Suppose an object is launched from ground level directly upward at 57.4 f/s Write a function to represent the object’s height ov

Semmy [17]

Answer: p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

Step-by-step explanation:

We can suppose that the only force acting on the object is the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration.

Then we can write:

a(t) = -32 ft/s^2

Where the negative sign is because this acceleration is downwards.

Now, to get the vertical velocity of the object, we need to integrate over time to get:

v(t) = (-32 ft/s^2)*t + v0

where t represents time in seconds and v0 is the constant of integration, and in this case, is the initial vertical velocity.

In this case, the initial velocity is 57.4 ft/s upwards, then the velocity equation is:

v(t) = (-32 ft/s^2)*t + 57.4 ft/s

To get the position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t + p0

Where p0 is the initial height of the object, as it was launched from the ground, then the initial position is p0 = 0ft.

then the position equation (that is the function that represents the height of the object as a function over time) is:

p(t) = (1/2)*(-32 ft/s^2)*t^2 + (57.4 ft/s)*t

p(t) = (-16 ft/s^2)*t^2 + (57.4 ft/s)*t

3 0

3 years ago

Which inequality is true?

11Alexandr11 [23.1K]

Answer:

$9\pi >27$

Step-by-step explanation:

$9\pi >27$

Divide both sides by 9.

$9\pi/9 >27/9$

$\pi >3$

$3.14159265358.. >3$

True.

7 0

3 years ago