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3 years ago

There is a line whose slope is 10 and whose y -intercept is 6. What is its equation in slope -intercept form?

Mathematics

2 answers:

Studentka2010 [4]3 years ago

6 0

Answer:

The equation is y = 10x + 6.

Step-by-step explanation:

In a slope-form equation, y = mx + b, m represents gradient and b is y-intercept. Given that slope is 10 and y-intercept is 6. So the equation will be y = 10x + 6.

Bingel [31]3 years ago

5 0

Answer:

y = 10x + 6

Step-by-step explanation:

slope-intercept (a.k.a. y-intercept) form is y = mx + b

m represents the slope and b represents the y intercept, so when writing in slope-intercept form, just replace m and b with that information.

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Which equation describes the circle having center point (3,7) and the radius r=4 in Standard form

Annette [7]

The equation of circle in standard form is $(x - 3)^2 + (y - 7)^2 = 16$

<h3><u>Solution:</u></h3>

Given that circle having center point (3,7) and the radius r = 4

To find: equation of circle in standard form

<em><u>The equation of circle is given as:</u></em>

$(x - h)^2 + (y - k)^2 = r^2$

Where center (h,k) and radius r units

Given that center point (h , k) = (3, 7) and radius r = 4 units

Substituting the values in above equation of circle,

$(x - 3)^2 + (y - 7)^2 = 4^2\\\\(x - 3)^2 + (y - 7)^2 = 16$

Thus the equation of circle in standard form is $(x - 3)^2 + (y - 7)^2 = 16$

6 0

3 years ago

Dentify the type of variation given and the constant of variation (k)

velikii [3]

The constant variation k is 12

3 0

3 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$