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3 years ago

13

What is the combined weight of 0.5 pounds bag?

1 answer:

kvasek [131]3 years ago

3 0

Answer:

7lbs

Step-by-step explanation:

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Help me please I have 10 minutes to do this

Olin [163]

Answer:

Distributive

Step-by-step explanation:

The 4y is distributed to the terms in parentheses.

7 0

3 years ago

T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is

K(x)= $\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}$

The given curve is y=7 $e^{x}$

${y}''=7e^{x}\\ {y}'=7e^{x}$

k(x)= $\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

${k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}$

For Maxima or Minima

${k(x)}'=0$

$7(e^x)(1+49e^{2x})(98e^{2x}-1)=0$

→ $e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0$

$e^{x}=0 , ∧ 1+49e^{2x}=0$ [not possible ∵there exists no value of x satisfying these equation]

→ $98e^{2x}-1=0$

Solving this we get

x= $-\frac{1}{2}\ln{98}$

As you will evaluate ${k(x})}''$ <0 at x= $-\frac{1}{2}\ln98$

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[ $-\frac{1}{2}\ln98$ ,1/√2]

k(x)= $\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

k(x)= $\frac{7}{\infty}$

k(x)=0

5 0

3 years ago

What is the product of (3 the square root of 8)(4 the square root of 3)? Simplify your answer

Vsevolod [243]

$3\sqrt{8}\cdot4\sqrt{3} \\12\sqrt{24} \\12\cdot2\sqrt{6} \\\boxed{24\sqrt{6}}$

8 0

3 years ago

Identify any solutions to the system shown here.

OLga [1]

The answer would be B and D. This can be found by putting the x and y values into the inequality to find whether or not they are true.

8 0

3 years ago

Read 2 more answers

HURRY!!!!!!!!!<br><br> Given P(5,10,8) and Q(7,11,10), find the midpoint of the segment PQ.

rusak2 [61]

Answer:

$P_{m}=(6,10.5,9)$

Step-by-step explanation:

The mid point can be found with the formula

$P_{m}=(\frac{x_{1}+x_{2} }{2},\frac{y_{1} +y_{2} }{2} ,\frac{z_{1}+z_{2} }{2} )$

The given coordinates are $P(5,10,8)$ and $Q(7,11,10)$ .

Replacing coordinates in the formula, we have

$P_{m}=(\frac{5+7}{2},\frac{10+11 }{2} ,\frac{8+10}{2} )=(\frac{12}{2},\frac{21 }{2} ,\frac{18}{2} )\\P_{m}=(6,10.5,9)$

Therefore, the mid point of the segment PQ is $P_{m}=(6,10.5,9)$

4 0

3 years ago