The solution to 4p+2<2(p+5) is

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer:

p<4

Step-by-step explanation:

4p+2< 2(p+5)

Distribute: 4p+2<2p+10

Subtract 2 from both sides: 4p<2p+8

Subtract 2p from both sides: 2p<8

Divide and we get: p<4

You might be interested in

Please Help! In the heavyweight class of professional wrestling, the junior weight limit is 190 pounds. This is 15 pounds heavie

lina2011 [118]

190
- 15
------
175 pounds

7 0

3 years ago

Helppppp plzzzzz 10points

wel

I think it should be B because the depth of the water should be decreasing slowing at the same rate.

4 0

3 years ago

Read 2 more answers

Rewrite in Polar Form....<br> x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

3 years ago

How much will he spend if he buys 4 pizzas and 5 drinks?

gtnhenbr [62]

Answer:

How much is each pizza and how much is each drink? set up your equation as 4p+5d = (your answer)

Step-by-step explanation:

3 0

3 years ago

Two plus ten times eight

alexdok [17]

Two plus ten times eight
(2+10)*8 = 96

3 0

3 years ago