Consider the system

$\{\begin{array}{l}1=A+B=C+D\\ B\ge C\end{array}$

with $A,B,C,D$ positive.

Does the system imply that $A\le D$?

$\{\begin{array}{l}1=A+B=C+D\\ B\ge C\end{array}$

with $A,B,C,D$ positive.

Does the system imply that $A\le D$?

Jamison Rios
2022-07-13
Answered

Consider the system

$\{\begin{array}{l}1=A+B=C+D\\ B\ge C\end{array}$

with $A,B,C,D$ positive.

Does the system imply that $A\le D$?

$\{\begin{array}{l}1=A+B=C+D\\ B\ge C\end{array}$

with $A,B,C,D$ positive.

Does the system imply that $A\le D$?

You can still ask an expert for help

asked 2022-05-23

Graph the following system of inequalities. Show (by shading in) the feasible region.

$x+2y\le 12$

$2x+y\le 12$

$x\ge 0,y\ge 0$

$x+2y\le 12$

$2x+y\le 12$

$x\ge 0,y\ge 0$

asked 2022-07-07

Suppose we are given a rational $q$. Is it possible that there infinitely many integer solutions $(h,k)$ to following system of inequalities: $0<|q-h/k|<1/2{k}^{2}$? I think that it is easy to see that for $q=1/2$ (and probably for any rational $a/b$ with $2|b$ and $gcd(a,b)$$=1$) it is impossible, but what about other rationals? What about changing constant in the right inequality (making it stronger or weaker)?

asked 2021-02-25

Graph the solution set for the system of linear inequalities

asked 2022-06-30

SAT inequality problem

$y\le 3x+1$

$x-y>1$

Which of the following ordered pairs $(x,y)$ satisfies the system of inequalities above?

$A)(-2,-1)$

$B)(-1,3)$

$C)(1,5)$

$D)(2,-1)$

$y\le 3x+1$

$x-y>1$

Which of the following ordered pairs $(x,y)$ satisfies the system of inequalities above?

$A)(-2,-1)$

$B)(-1,3)$

$C)(1,5)$

$D)(2,-1)$

asked 2022-05-17

Are there an infinite number of prime quadruples of the form $10n+1$, $10n+3$, $10n+7$, $10n+9$?

In base $10$, any prime number greater than 5 must end with the digits 1$1$, $3$, $7$, or $9$. For some $n$, $10n+1$, $10n+3$, $10n+7$, $10n+9$ are all prime: for example, when $n=1$, we have that $11$, $13$, $17$, and $19$ are all prime. My question is, can anyone disprove the claim that there are an infinite number of such primes.

The only progress I've been able to make is to show that $n$ must be of the form $3k+1$ by considering the system of modular inequalities

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2$

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3$

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}5$

In base $10$, any prime number greater than 5 must end with the digits 1$1$, $3$, $7$, or $9$. For some $n$, $10n+1$, $10n+3$, $10n+7$, $10n+9$ are all prime: for example, when $n=1$, we have that $11$, $13$, $17$, and $19$ are all prime. My question is, can anyone disprove the claim that there are an infinite number of such primes.

The only progress I've been able to make is to show that $n$ must be of the form $3k+1$ by considering the system of modular inequalities

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2$

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}3$

$p\not\equiv 0\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}5$

asked 2022-06-20

Let $u,v:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ satsfying the following system of differential inequalities:

${u}^{\prime}(t)\le {a}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{a}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{a}_{0}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{\prime}(t)\le {b}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{b}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{b}_{0}$

for suitable coefficients ${a}_{0},{a}_{1},{a}_{2},{b}_{0},{b}_{1},{b}_{2}\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}.$ In particular I have ${a}_{1},{b}_{2}<0\phantom{\rule{thinmathspace}{0ex}}$ and $0<{b}_{1}<|{b}_{2}|\phantom{\rule{thinmathspace}{0ex}}$

Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type

$u(t)\le F(t)\phantom{\rule{0ex}{0ex}}v(t)\le G(t)$

where the functions $F,G:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ depend on $u,v$ only through their initial values $u(0),v(0)$?

${u}^{\prime}(t)\le {a}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{a}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{a}_{0}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{\prime}(t)\le {b}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{b}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{b}_{0}$

for suitable coefficients ${a}_{0},{a}_{1},{a}_{2},{b}_{0},{b}_{1},{b}_{2}\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}.$ In particular I have ${a}_{1},{b}_{2}<0\phantom{\rule{thinmathspace}{0ex}}$ and $0<{b}_{1}<|{b}_{2}|\phantom{\rule{thinmathspace}{0ex}}$

Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type

$u(t)\le F(t)\phantom{\rule{0ex}{0ex}}v(t)\le G(t)$

where the functions $F,G:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ depend on $u,v$ only through their initial values $u(0),v(0)$?

asked 2022-07-25

Graph the rational function: f(x) = $\frac{2{x}^{2}}{{x}^{2}-1}$ andgive the following information:

Domain:

y intercept:

Horizontal asymptote:

Vertical Asymptote(s):

x intercept(s):

Complete the Table:

x f(x)

-3

-2

-0.5

0.5

2

3

(b) using the answer from (A), solve the rationalequality:

$\frac{2{x}^{2}}{{x}^{2}-1}<0$

Domain:

y intercept:

Horizontal asymptote:

Vertical Asymptote(s):

x intercept(s):

Complete the Table:

x f(x)

-3

-2

-0.5

0.5

2

3

(b) using the answer from (A), solve the rationalequality:

$\frac{2{x}^{2}}{{x}^{2}-1}<0$