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Igoryamba
3 years ago
5

Solve m/26=-1/2 '''''''''''''''''''''''

Mathematics
1 answer:
trapecia [35]3 years ago
6 0

Answer:

m= -13

Step-by-step explanation:

step1:  left part times 26 is   m

step2: right part times 26 is -13

so m= -13

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A school cafeteria sold 1,280 slices of pizza the first week,640 the second week,and 320 the third week.if this pattern continue
mario62 [17]
So we have 1280,640,320,... 
This is a geometric sequence with the first term, a_{1} =1280. To find the common ratio r, we are going to divide any current term by a previous one: r= \frac{640}{1280} =(0.5)

Remember that the main formula of a geometric sequence is:
a_{n} = a_{1} r^{n-1}
Where a_{n} is the nth term (in our case 40), a_{1} is the first term (in our case 1280), r is the common ratio (0.5), and n is the position of the term in the sequence (in our case our weeks)

Now we can replace the values to get:
40=1280(0.5)^{n-1}
(0.5)^{n-1} = \frac{40}{1280}
(0.5)^{n-1} =0.03125
Since our variable, n, is the exponent, we are going to use logarithms to bring it down:
ln(0.5)^{n-1} =ln(0.03125)
(n-1)ln(0.5)=ln(0.03125)
The only thing left now is solving for n to find our week:
n-1= \frac{ln(0.03125)}{ln(0.5)}
n-1=5
n=6

We can conclude that in the sixth week the cafeteria will sell 40 slices of pizza.
5 0
3 years ago
Ms.smith has 132 stickers to give to her students. she has 25 students. if she gives the same number of stickers to each of her
zhuklara [117]

Answer:

Step-by-step explanation:

If Ms. Smith hands out 130 of her stickers between all 5 of her students each will have 26. Which means she will have 2 remaining

5 0
4 years ago
Evaluate : limx→ tan2-sin2x x3
GrogVix [38]

Given:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Solve:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Use l'hopital's rule:

\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}

Simplify:

\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}

Apply the constant multiple rule:

\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}

Similary :

\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}

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Answer:

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3 years ago
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Find the area of the rectangle
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