Please help its inequalities 2x

Find the perimeter of a triangle with sides lengths 8, 10, and 12.

shtirl [24]

Answer:

30

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please help me plsssssss! asappp. I'll mark brainlest and give points

Nookie1986 [14]

Answer:

Step-by-step explanation:

1)a) Abscissa of O 0

Abscissa of A 4

Abscissa of B -3 {opposite of 3}

Abscissa of C 3 {3 is the only positive integer like A}

Abscissa of D -4.5

Abscissa of E -6

Abscissa of F -1 {Midpoint of AE = (6+4)/2 = 10/2 = 5th number from -6 or 4}

b) OB = 3 units

DA = 8.5 {4.5 +4 = 8.5}

7 0

2 years ago

I wasn't in class so I need help solving the inequality 7/10n+14<49

NikAS [45]

Answer:

n<50

Step-by-step explanation:

7/2*5n + 14<49

7n/2*5+14<49

(7n)+(2*5)14/2*5 <49

7n+10*14/2*5 <49

7n+140/2*5 <49

7n+140/10 <49

7n+140 < 10*49

7n+140 < 490

(7n+140)+(-140)<490+(-140)

7n+140-140<490-140

7n<350

7n/7 < 350/7

n<2*5^2*7/7

n<2*5^2

n<2*25

n<50

8 0

2 years ago

the sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger

Reil [10]

<h2>Answer:</h2>

$x\geq 3 \ and \ x+1 \geq 4$

<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable $x$ which is the smaller number. Accordingly, the lager number will be $x+1$ given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

$\leq \\ which \ means \ less \ than$

So:

1. The sum of 2 consecutive integers can be written as:

$v+(v+1)$

2. Nine times the smaller and 5 times the larger can be written as:

$9v-5(v+1)$

Finally, the whole statement:

The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\$

$x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\ \frac{6}{2} \leq \frac{2x}{2} \\ \\ 3 \leq x \\ \\ x\geq 3 \\ \\ and \\ \\ x+1 \geq 4$

The two numbers are:

$x\geq 3 \ and \ x+1 \geq 4$

6 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

a.

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

2 years ago

Please help its inequalities 2x < 6