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defon
3 years ago
9

6th grade math i mark as brainliest​

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

Answer:

30

Step-by-step explanation:

I think that's the answer

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30 pts MATH PROBLEM AGAIN!! A trick coin is designed to land heads up with a probability of 80%
goldenfox [79]
I think d is the answer
8 0
4 years ago
Write an equation in slope intercept form of the line with the given characteristics:
stiks02 [169]

Answer:

The equation of line through (-3,1) and slope 2 in slope-intercept form is: y = 2x+7

Step-by-step explanation:

The slope-intercept form of an equation is given by:

y = mx+b

here m is the slope of the line and b is the y-intercept

Given

m = 2

Putting m=2 in the general form

y = 2x+b

In order to find the y-intercept, the given point (-3,1) will be substituted in the equation

So

1 = 2(-3)+b\\1 = -6+b\\b = 1+6\\b = 7

Putting the value of b

y = 2x+7

Hence,

The equation of line through (-3,1) and slope 2 in slope-intercept form is: y = 2x+7

5 0
3 years ago
PLEASE PLEASE HELP ME WITH THIS!
andreev551 [17]

No. You can actually solve this question without doing any math. Her only score that is above a 90 is 93, so it will be brought down lower than a 90 very easily by the 77 and the 82.

5 0
4 years ago
What is the equation in standard form or the line that passes through the point ( 4,-8 ) and has a slope of 1/4?
ehidna [41]
The equation in y=mx+b form is 1/4x-9.
3 0
3 years ago
Water flows into a right cylindrical shaped swimming pool with a circular base at a rate of 4 m33/min. The radius of the base is
Sloan [31]

Answer:

The water level is rising at a rate of approximately 0.1415 meters per minute.

Step-by-step explanation:

Water is flowing into a right cylindrical-shaped swimming pool at a rate of 4 cubic meters per minute. The radius of the base is 3 meters.

And we want to determine the rate at which the water level of the pool is rising.

Recall that the volume of a cylinder is given by:

\displaystyle V = \pi r^ 2h

Since the radius is a constant 3 meters:

\displaystyle V = 9\pi h

Water is flowing at a rate of 4 cubic meters per minute. In other words, dV/dt = 4 m³ / min.

Take the derivative of both sides with respect to <em>t: </em>

<em />\displaystyle \frac{d}{dt}\left[ V\right] = \frac{d}{dt}\left[ 9\pi h\right]<em />

Implicitly differentiate:

\displaystyle \frac{dV}{dt} = 9\pi \frac{dh}{dt}

The rate at which the water level is rising is represented by dh/dt. Substitute and solve:

\displaystyle \left(4 \right) = 9\pi \frac{dh}{dt}

Therefore:

\displaystyle \frac{dh}{dt} = \frac{4}{9\pi} \approx 0.1415\text{ m/min}

In conclusion, the water level is rising at a rate of approximately 0.1415 meters per minute.

6 0
3 years ago
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