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Dahasolnce [82]
3 years ago
7

Select all that apply.

Mathematics
1 answer:
alexgriva [62]3 years ago
5 0
X+1
I hope that right
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Find the 6th term of the arithmetic sequence X – 9, -X – 11, -3.3 – 13, ...
MrMuchimi

Answer:

-13x-23

Step-by-step explanation:

Given sequence:  x-9, -x-11, -3x-13

Therefore,

  • a_1=x-9
  • a_2=-x-11
  • a_3=-3x-13

General form of an arithmetic sequence: a_n=a+(n-1)d

(where a is the first term and d is the common difference)

To find the common difference, subtract a term from the next term:

\begin{aligned}d & =a_2-a_1\\ & =(-x-11)-(x-9)\\ & = -x-11-x+9\\ & = -2x-2\end{aligned}

Therefore,

a_n & =(x-9)+(n-1)(-2x-2)

To find the 6th term, input n = 6 into the equation:

\begin{aligned}\implies a_6 & =(x-9)+(6-1)(-2x-2)\\ & = (x-9)+7(-2x-2)\\ & = x-9-14x-14\\ & = -13x-23\end{aligned}

3 0
2 years ago
200 labourers need 50 days to construct a road of 4 km long. How many labourers are needed to construct 8 km long in 80 days​
patriot [66]

Answer:

250

Step-by-step explanation:

Number of labourers * Number of days worked = Total number of working days

Total number of working days to construct a 4km long road: 200*50=10000

Total mumber of working days to construct a 8km long road: 10000*2=20000

Labourers: 20000 ÷ 80 = 250.

250 labourers are required.

7 0
3 years ago
One of the great mysteries of the world is my hotdogs and hotdog buns come in packages of different sizes hotdogs come in packs
V125BC [204]

Answer:

3 pack of hot dogs and 2 pack of buns 24

Step-by-step explanation:

hot dogs come in packs of 8

burns came in packs of 12

what is the least number of each you can buy in order to have a equal number of hot dogs and buns

so to get equal number

we have to look for the LCM of 8 and 12 = 24

therefore the packs will be equal to product

hot dogs come in packs of 8 = 8*3 = 24

burns came in packs of 12 = 12 *2 = 24

therefore, 3 pack of hot dogs and 2 pack of buns 24

7 0
3 years ago
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

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Display

6 0
3 years ago
What is the percent on my quiz if i got 32/44
kati45 [8]
You got a 72%! sorry! It's still a C!
8 0
3 years ago
Read 2 more answers
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