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3 years ago

What is 7+|−9−2x|=18

Mathematics

2 answers:

miss Akunina [59]3 years ago

8 0

Step-by-step explanation:

7+(-9-2x)=18

-9-2x=18-7

-9-2x=11

-2x=11+9

_2x=20

x=20

-2

x= -10

hope u want this..

Hitman42 [59]3 years ago

5 0

Answer Step by Step:

Move all terms not containing |-9-2x| to the right side of the equation.

|-9-2x| = 11

Remove the absolute value term. This creates a ± on the right side of the equation because |x| = ±x.

-9-2x=±11

Set up the positive portion of the ± solution.

-9-2x=11

Solve the first equation for x.

x = -10

Set up the negative portion of the ± solution.

-9-2x=-11

Solve the second equation for x.

x = 1

(Answer) x = -10,1

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given the recursive formula for a geometric sequence find the common ratio the 8th term and the explicit formula.did I set these

lesya [120]

Answer:

Step-by-step explanation:

1)Since we know that recursive formula of the geometric sequence is

$a_{n}=a_{n-1}*r$

so comparing it with the given recursive formula $a_{n}=a_{n-1}*-4$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-2*(-4)^{7} =32768.$

Explicit Formula = $-2*(-4)^{n-1}$

2) Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=-4*(-2)^{7} =512.$

Explicit Formula = $-4*(-2)^{n-1}$

3)Comparing the given recursive formula $a_{n}=a_{n-1}*3$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =3

8th term= $a_{1}*(r)^{n-1}=-1*(3)^{7} =-2187.$

Explicit Formula = $-1*(3)^{n-1}$

4)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=3*(-4)^{7} =-49152.$

Explicit Formula = $3*(-4)^{n-1}$

5)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-4*(-4)^{7} =65536.$

Explicit Formula = $-4*(-4)^{n-1}$

6)Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=3*(-2)^{7} =-384.$

Explicit Formula = $3*(-2)^{n-1}$

7)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=4*(-5)^{7} =-312500.$

Explicit Formula = $4*(-5)^{n-1}$

8)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=2*(-5)^{7} =-156250.$

Explicit Formula = $2*(-5)^{n-1}$

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3 years ago

Another question about angles

zvonat [6]

Answer:

ㄥBEC=70；ㄥABE=160

Step-by-step explanation:

(x-5)+(3x-5)+90=180

x-5+3x-5+90=180

4x+80=180

4x=100

x=25

ㄥBEC=3*25-5=70

ㄥABE=180-(25-5)=160

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vovikov84 [41]

Answer: 0.3439

Step-by-step explanation:

Given :The last four digits for telephone numbers are randomly selected (with replacement).

Here , each position can be occupied with any of the digit independently .

Total digits = 10

Total digits other than 0 = 9

For each digits , the probability that it is not 0 = $\dfrac{9}{10}=0.9$

If we select 4 digits , The probability of getting no 0 = $(0.9)^4 =0.6561$

(By multiplication rule of independent events)

Now , the probability that for one such phone number, the last four digits include at least one 0. = 1- P(none of them is 0)

=1- 0.6561=0.3439

Hence, the probability that for one such phone number, the last four digits include at least one 0. is 0.3439 .

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