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2 years ago

15

Round the number 15.39624 to 2 decimal places

1 answer:

ivolga24 [154]2 years ago

7 0

Answer:

answer is 15.40 because 9 is greater than 5 so 1 is carry over it .So answer is 15.40

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Which of the following is the graph of (x-2)^2 + (y+1)^2 = 9? (Brainly doesn't seem to want to let me put up pictures. Sorry guy

forsale [732]

If you want to cheat, paste the formula in Wolfram Alpha. However, this one is not that difficult. From the fact that x and y are squared you can infer that this is a circle. The 9 reveals that the radius r^2 is 9, so radius is 3.

The (x-2) tells us that the circle is shifted to the right by 2. Likewise, y+1 shifts it down by one. So the center is at (2,-1). If the equation would be x^2 + y^2 = 9, you'd have a circle exactly on the origin.

Hopefully this helps you break down the equation and pick the right picture!

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3 years ago

Whats the answer and equation to Two more than four times a number is -10

Artist 52 [7]

2+4a=-10

First, subtract 2 from both sides.

4a=-12

Then divide 4 from both sides.

a=-3

hope it helps!

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3 years ago

What is the slope of the line y = -5 - 3x ?<br> o 3<br> O 5<br> 0 -3<br> 0 -8

nexus9112 [7]

Answer:

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Step-by-step explanation:

but, its 0 -8

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3 years ago

What is the slope and equation of a line parallel to the x-axis passing through point (3,7)

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6 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

2 years ago