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3 years ago

15

If 5^2x+1 = 96, Then find x

1 answer:

Mkey [24]3 years ago

6 0

Jkeoeodiekqnsnabbdbw

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Solve. the square root of 44-2x = x-10

Anuta_ua [19.1K]

44-2x=x-10 subtract 44 from both sides -2x= x-54 subtract x from both sides -3x= -54 divide by -3 x= 18 square root of 18= 4.24 in radical form 3 square root of 2 hope this helps

5 0

3 years ago

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Pls Help i dont understand ittttttttttt

agasfer [191]

The given scale for the map means that a distance of 1 cm between two locations on the map corresponds to a real distance between them of 600,000 cm = 6 km.

So, if the distance between two places in reality is 87 km, on the map this would be represented by a distance of 14.5 cm, since

87 = 84 + 3

87 = 6 • 14 + 6 • (1/2)

87 = 6 • (14 + 1/2)

3 0

2 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

I giveeeeeeeeeeee brainlillstttttttttttt

Lana71 [14]

Answer:

after reflection over y axis

a 9,7

b 9,6

c 0,6

d 0,7

4 0

3 years ago

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Hi! Answer this please!

Rudiy27

The answer would be b

4 0

2 years ago

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