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2 years ago

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Consider this system of equations. Which shows the second equation written in slope-intercept form? y = 3 x minus 2. 10 (x + thr

ee-fifths) = 2 y y = 5 x + StartFraction 3 Over 10 EndFraction y = 5 x + 3 y = one-fifth x + StartFraction 3 Over 25 EndFraction y = one-half x + 6

2 answers:

Bas_tet [7]2 years ago

8 0

Answer:

The answer is B

Step-by-step explanation:

Can i have Brainliest

Arte-miy333 [17]2 years ago

7 0

Question

Consider this system of equations. Which shows the second equation written in slope-intercept form?

$y = 3x - 2.$

$10(x + \frac{3}{5} ) = 2y$

A. $y = 5x + \frac{3}{10}$

B. $y = 5x + 3$

C. $y = \frac{1}{5} x + \frac{3}{25}$

D. $y = \frac{1}{2} x + 6$

Answer:

B. $y = 5x + 3$

Step-by-step explanation:

Given

Equation 1: $y = 3x - 2.$

Equation 2: $10(x + \frac{3}{5} ) = 2y$

Required:

Equivalent of equation 2

To get an equivalent of equation 2 (in slope intercept form), first we have to simplify equation 2

$10(x + \frac{3}{5} ) = 2y$

Open the bracket

$10*x + 10 *\frac{3}{5} = 2y$

$10x + \frac{30}{5} = 2y$

Simplify fraction

$10x + 6 = 2y$

Divide through by 2

$\frac{10x}{2} + \frac{6}{2} = \frac{2y}{2}$

$5x + 3 = y$

Re-arrange

$y = 5x + 3$

The next step is to compare each of option A through D with $y = 5x + 3$

A.

$y = 5x + \frac{3}{10}$ is not equal to $y = 5x + 3$

We check the next available option

B.

$y = 5x + 3$ is equal to $y = 5x + 3$

This option is equivalent to the second equation in slope-intercept form.

We check further if there are more equivalent options

C.

$y = \frac{1}{5} x + \frac{3}{25}$

Convert fraction to decimal

$y = 0.2x + 0.12$

This is not equal to $y = 5x + 3$

D.

$y = \frac{1}{2} x + 6$

Convert fraction to decimal

$y = 0.5x + 6$

This is not equal to $y = 5x + 3$

Hence, the only equation that is equivalent to the second equation written in slope intercept form is Option B

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2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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2 years ago

A company manufactures a 14-ounce box of cereal. Boxes are randomly weighed to ensure the correct amount. If the discrepancy in

Stolb23 [73]

Answer:

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Step-by-step explanation:

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2 years ago

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Ilya [14]

Answer:

Value of expression in single logarithm is $\log_3\left(2z^2\right)$ .

Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

$\log_3\left(\left(64z^2\right)\right)-\log_3\left(3^4-7^2\right)$

Applying quotient rule of logarithmic,

$\log_c\left(a\right)-\log_c\left(b\right)=\log_c\left(\frac{a}{b}\right)$

Therefore,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{3^4-7^2}\right)$

Simplifying,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{81-49}\right)$

$\log_3\left(\dfrac{\left(64z^2\right)^2}{32}\right)$

$\log_3\left(2z^2\right)$

Therefore value of expression is $\log_3\left(2z^2\right)$

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2 years ago

Read 2 more answers

Convert the angle \theta=\dfrac{29\pi}{15}θ= 15 29π theta, equals, start fraction, 29, pi, divided by, 15, end fraction radian

iren [92.7K]

Answer:

348 degrees

Step-by-step explanation:

29 pi /15 to degrees

To change from radians to degrees, multiply by 180/pi

29 pi /15 * 180/pi

29 * 180/15

348

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2 years ago