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2 years ago

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Simplify:4(r + 1) + 3r A. 4r + 4 B. 5r + 1 C. 7r + 1 D. 7r + 4

2 answers:

klasskru [66]2 years ago

6 0

Answer:

look in the explanation, it is in there

Step-by-step explanation:

i don't know if you are distributing or just using PEMDAS .

if you are distributing then: do in the equation 4(r + 1) + 3r.. do 4 x r and 4 x 1 , now you have 4r + 4 + 3r

7r +4

is what i got!

If you are doing PEMDAS then: first you do what is inside the perenthisis ()

4(r + 1) +3r changes into

4( r1 ) + 3r now you do multipulcation because that in the "M" in pemdas

4r + 3r, then if you are using like terms you would add them

7r

i hope this helps you out!

scoray [572]2 years ago

5 0

The correct answer is , D. 7r + 4
using the pemdas method will result in the correct answer .

1. distribute 4 to r and 1
•4r and 4 (4r + 4)
2. combine like terms
•4r and 3r (4r+3r)
3. add like terms together
•4r + 3r = 7r
4. since our constant from step one has no other like terms , it stays as is resulting in our answer being : D. 7r + 4

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2 years ago

4/5 ÷ 7/6 + 2/7 ∙2 1/5

Aleksandr [31]

<h2>The value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ $=\dfrac{8}{7}$ </h2>

Step-by-step explanation:

We have,

$\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$

To find, the value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ = ?

∴ $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$

$=\dfrac{4}{5}\times \dfrac{6}{7} +\dfrac{2}{7}.\dfrac{8}{5}$

$=\dfrac{24}{35} +\dfrac{16}{35}$

Taking LCM of denominator, we get

$=\dfrac{24+16}{35}$

$=\dfrac{40}{35}$

Dividing numerator and denominator by 5, we get

$=\dfrac{8}{7}$

∴ The value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ $=\dfrac{8}{7}$

Hence, the value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ is equal to $\dfrac{8}{7} .$

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3 years ago

Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notat

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In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

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$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

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Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t)$

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For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

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For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$

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