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3 years ago

13

NEED AN ANSWER SOON

2 answers:

Zina [86]3 years ago

4 0

Answer:

3/2

Step-by-step explanation:

Yakvenalex [24]3 years ago

4 0

Answer : 3/2

To find the slope you would use the slope formula which is (y2-y1) / (x2-x1).

(8-2) / (7-3)

m = 3/2

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Which expression is equivalent to(8w 7x-5y3z-9)-2/3

Shkiper50 [21]

Answer:

$56wx - 15yz - \frac{29}{3}$

Step-by-step explanation:

$1. \: 56wx - 5y \times 3z - 9 - \frac{2}{3} \\ 2. \: 56wx - 15yz - 9 - \frac{2}{3} \\ 3. \: 56wx - 15yz + ( - 9 - \frac{2}{3} )$

4 0

3 years ago

With a short time remaining in the day, a delivery driver has time to make deliveries at 5 locations among the 9 locations remai

Delvig [45]

Answer:

15,120

Step-by-step explanation:

The computation of the number of different routes are possible is shown below:

Since there are 5 locations for deliveries among the 9 locations that left

So, the number of routes that could be possible is

$= ^9P_5\\\\= \frac{9!}{9-5!} \\\\= \frac{9!}{4!} \\\\= \frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}\\\\= 15,120$

5 0

3 years ago

Let x = -3, y = -2, and z= 7. What is the<br>value of this expression?<br>yz+ 2(2-x)

Alja [10]

Answer:

so the expression should be -2+7+2(2-(-3))

3 0

3 years ago

Can someone pls help me with this

Misha Larkins [42]

Can I see the whole question pls

5 0

3 years ago

Find the solution of the differential equation that satisfies the given initial condition. (dP)/(dt)

kati45 [8]

Answer:

$P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2$

Step-by-step explanation:

Given

$\frac{dP}{dt} = \sqrt{Pt$

$P(1) = 2$

Required

The solution

We have:

$\frac{dP}{dt} = \sqrt{Pt$

$\frac{dP}{dt} = (Pt)^\frac{1}{2}$

Split

$\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}$

Divide both sides by $P^\frac{1}{2}$

$\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}$

Multiply both sides by dt

$\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt$

Integrate

$\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt$

Rewrite as:

$\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt$

Integrate the left hand side

$\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt$

$\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt$

$2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt$

Integrate the right hand side

$2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c$

$2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c$

$2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c$ ---- (1)

To solve for c, we first make c the subject

$c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}$

$P(1) = 2$ means

$t = 1; P =2$

So:

$c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}$

$c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1$

$c = 2\sqrt 2 - \frac{2}{3}$

So, we have:

$2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c$

$2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}$

Divide through by 2

$P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}$

Square both sides

$P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2$

6 0

3 years ago