WILL GIVE BRAINLIEST!!!!

7 divided by 982 long division

Ne4ueva [31]

Answer:

140r2

Step-by-step explanation:

7 0

2 years ago

What is the slope to this line?

Artyom0805 [142]

Answer:

hello :

Hello : let A(1,0) B(0,2) in this line

the slope is : (YB - YA)/(XB -XA)

5 0

2 years ago

Write the equation of the line that has a slope of -2 and goes through the point (-3,4).

liubo4ka [24]

<u>SOLUTION:</u>

General Equation: y = mx + b

.

Substitute slope = -2 into the equation:

y = -2x + b

.

Find the y-intercept:

4 = -2(-3) + b

4 = 6 + b

b = -2

.

Substitute b = -2 into the equation:

y = -2x - 2

.

Answer: y = -2x - 2

4 0

3 years ago

Read 2 more answers

The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given

slavikrds [6]

Answer:

a) $\frac{dm}{dt} = -k\cdot m$ , b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ , c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$ , d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$

4 0

2 years ago

Read 2 more answers

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

sergeinik [125]

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$ , so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.

5 0

2 years ago