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3 years ago

Will give brainliest!

1 answer:

8 0

Answer is C. y=1/3x+3
the slope is going to be 1/3 (opposite of -3/1) and if go to the point 3,4 and go down one and left 3, you will get a y intercept as 3

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1. When a cannonball is fired, the equation of its pathway can be modeled by h = -1612 + 128t.

sashaice [31]

Answer:

The maximum height of the ball is 256m

Step-by-step explanation:

Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t

At maximum height, the velocity of the ball is zero.

velocity = dh/dt

velocity = -32t + 128

Since v = 0 at maximum height

0 = -32t+128

32t = 128

t = 128/32

t = 4seconds

The maximum height can be gotten by substituting t = 4 into the modelled equation.

h = -16t² + 128t

h = -16(4)²+128(4)

h = -16(16)+512

h = -256+512

h = 256m

4 0

3 years ago

Graph y=−4/7x+1. do this math

Fofino [41]

Answer:

Use demos or a graphing calc.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How to do this problem??

lorasvet [3.4K]

We know that the sum of the inner angles of any triangle is 180º

72º + (7x + 3)º + (3x + 5)º = 180º
72º + 7xº + 3º + 3xº + 5º = 180
7xº + 3xº = 180º - 72º - 3º - 5º
10xº = 100º
$x^0 = \frac{100^0}{10^0}$
$x^0 = 10\to\:\boxed{\boxed{x = 10^0}}\end{array}}\qquad\quad\checkmark$

The sum of the external angle (9y + 1)º with inner angle (3x + 5) = 180 °, <span>Replace the measure of "x" found:
</span>
(9y + 1)º + (3x + 5)º = 180º
9yº + 1º + 3xº + 5º = 180º
9yº + 1º + 3.(10)º + 5º = 180º
9yº + 1º + 30º + 5º = 180º
9yº = 180º - 1º - 30º - 5º
9yº = 144º
$y^0 = \frac{144^0}{9^0}$
$y^0 = 16\to\:\boxed{\boxed{y = 16^0}}\end{array}}\qquad\quad\checkmark$

Answer:
<span>The measures of "x" and "y" are respectively: 10º and 16º</span>

6 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$