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3 years ago

Please help me I’ll give you brainless it’s due in 20 minutes

Mathematics

1 answer:

ankoles [38]3 years ago

7 0

Answer:

*In Explanation

Step-by-step explanation:

1) Start by just substituting the value of x (given in equation 1) into the second equation:

y = 3x - 1

y = 3(2) - 1

y = 6-1 = 5

2) Since both equations have x on one side, make both equations equal to one another:

2y + 4 = 9 - 3y

Solve for y:

5y = 5

y = 1

Plug y = 1 into either one of the given equations and solve for x:

I'll use first equation:

x = 2y + 4

x = 2(1) + 4

x = 6

3) When substituting X into the second equation, remember to use parenthesis:

The student was substituting x from equation 1 into equation 2, but they forgot to multiply ALL of 2y + 3 by 2.

x = 2y + 3

Substitute into 2nd eqn:

y = 2(2y + 3) - 9

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Number 1 is 32 and the second one is 85

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3 years ago

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

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Determine the width of the pond. XL. 90 ft. 135 ft. 120 ft

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4 years ago

A report on consumer financial literacy summarized data from a representative sample of 1,570 adult Americans. Based on data fro

alexgriva [62]

Answer:

a) 95% confidence interval for the proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance = (0.498, 0.547)

This means we are 95% confident that the true proportion all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is within the range 49.8% and 54.7%.

b) The confidence interval from part (a) is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B because the interval obtained contains proportions that are greater than 50% indicating that there is significant evidence that the true proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is more than half of the total population.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = (820/1570) = 0.5223

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 95% confidence interval for sample size of 1570 is obtained from the z-tables.

Critical value = 1.960

Standard error of the mean = σₓ = √[p(1-p)/n]

p = sample proportion = 0.5223

n = sample size = 1570

σₓ = √(0.5223×0.4777/1570) = 0.0126063049 = 0.01261

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.5223 ± (1.96 × 0.01261)

CI = 0.5223 ± 0.02471

95% CI = (0.4975916424, 0.5470083576)

95% Confidence interval = (0.4976, 0.5470)

We are 95% confident that the true proportion all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is within the range 49.8% and 54.7%.

Hope this Helps!!!!

7 0

3 years ago