All categories

3 years ago

Solve the inequality: 12p + 7 > 139

2 answers:

4 0

If you would like to solve the inequality 12 * p + 7 > 139, you can do this using the following steps:

12 * p + 7 > 139
12 * p > 139 - 7
12 * p > 132 /12
p > 132/12
p > 11

The correct result would be B. p > 11.

laila [671]3 years ago

4 0

12p + 7>139
12p>139-7
12p>132
p>132/12
p>11

B.p>11

You might be interested in

If k is a positive integer and n = k(k + 7), is n divisible by 6 ? (1) k is odd. (2) When k is divided by 3, the remainder is 2.

olganol [36]

Answer:

1.) Yes

2.) Yes

Step-by-step explanation:

Given that

n = k(k + 7)

If k is a positive integer and n = k(k + 7), is n divisible by 6 ?

(1) k is odd. Yes.

Let assume that k = 3

Then, n = 3(3 + 7)

n = 3 × 10

n = 30.

30 is divisible by 6.

(2) When k is divided by 3, the remainder is 2. That is,

Let k = 5

Then,

5/3 = 1 remainder 2

Substitute k into the equation

n = k(k + 7)

n = 5(5 + 7)

n = 5 × 12

n = 60

And 60 is divisible by 6.

Therefore, the answer to both questions is Yes.

5 0

3 years ago

A line passing through which of the following pairs of coordinates represents a proportional relationship? A. (4, 5) and (8, 7)

galina1969 [7]

B if you multiply x and y by 2 there you go!!

4 0

3 years ago

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

Point a:

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

Point b:

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for y we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$