Which of the following best describes the slope of the line below?

What is 5^(−3)×2^(−1)

Valentin [98]

The answer to this question is 0.004

3 0

3 years ago

How can finding the slopes of MN and AB help in proving that MN is the midsegment of triangle ABC?

Mademuasel [1]

Mid-segment and bottom side of the triangle is always parallel to each other, and parallel lines must have same slopes, so if their slopes are equal, then MN would be mid-segment of side AB in Triangle ABC

Hope this helps!

4 0

4 years ago

Read 2 more answers

Simplify: (ax+2)(ax+b)(b)0

Lina20 [59]

0 anything multiplied by Zero is zero

6 0

3 years ago

Read 2 more answers

If Lisa’s gross pay was $1204.89 and her net pay was $818.44, about what percent of her pay was deducted?

Amanda [17]

Answer:

About 32% of her pay was deducted ⇒ 2nd answer

Step-by-step explanation:

Lisa’s gross pay was $1204.89

Her net pay was $818.44

We need to find the percentage of the amount deducted from her pay

Lets calculate the deduction amount

The deduction amount is the difference between the gross pay and

the net pay

∵ The amount deducted = gross pay - net pay

∵ Gross pay = $1204.89

∵ Net pay = $818.44

∴ The amount deducted = 1204.89 - 818.44 = $386.45

Now lets find the percentage of the amount deducted

∵ Percent of deducted = [amount deducted/gross pay] × 100%

∵ The amount deducted = $386.45

∵ Gross pay = 1204.89

∴ Percent of deducted = $\frac{386.45}{1204.89}$ × 100%

∴ Percent of deducted = 32.07%

About 32% of her pay was deducted

8 0

3 years ago

Integrate <img src="https://tex.z-dn.net/?f=e%5E%7B4x%7D%5Csqrt%7B1%2Be%5E%7B2x%7D%20%7D%20dx" id="TexFormula1" title="e^{4x}\sq

AnnyKZ [126]

Answer:

$(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Step-by-step explanation:

Step(i):-

Given that the function

f(x) = $e^{4x} \sqrt{1+e^{2x} }$

Now integrating on both sides, we get

$\int\limits{f(x)} \, dx = \int\limits{e^{4x} \sqrt{1+e^{2x} } dx$

= $\int\limits{e^{2x} e^{2x} \sqrt{1+e^{2x} } dx$

Step(ii):-

Let $1 + e^{2x} = t$

$e^{2x} = t -1$

$2e^{2x}dx = d t$

$e^{2x}dx = \frac{1}{2} d t$

= $\int\limits{( \sqrt{1+e^{2x} }) e^{2x} e^{2x} dx$

= $\int\limits {\sqrt{t}(t-1)\frac{1}{2} dt }$

= $\frac{1}{2} \int\limits {\sqrt{t} (t) -\sqrt{t} ) dt }$

= $\frac{1}{2} \int\limits {(t^{\frac{1}{2} } t^{1} +t^{\frac{1}{2} } ) } \, dx$

$= \frac{1}{2} \int\limits {(t^{\frac{3}{2} } +t^{\frac{1}{2} } ) } \, dx$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{3}{2}+1 } + \frac{t^{\frac{1}{2} +1} }{\frac{1}{2}+1 } )+C$

= $\frac{1}{2} (\frac{t^{\frac{3}{2} +1} }{\frac{5}{2} } + \frac{t^{\frac{1}{2} +1} }{\frac{3}{2} } )+C$

= $\frac{1}{2} (\frac{t^{\frac{5}{2} } }{\frac{5}{2} } + \frac{t^{\frac{3}{2} } }{\frac{3}{2} } )+C$

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

Final answer:-

= $(\frac{(1+e^{2x}) ^{\frac{5}{2} } }{{5}} + \frac{(1+e^{2x} )^{\frac{3}{2} } }{{3}} )+C$

3 0

3 years ago