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2 years ago

13

Can somebody help solve either of these?

1 answer:

Leya [2.2K]2 years ago

8 0

First one:
The circumference of the whole circle is (2π x the radius),
and 60° is (60/360) = 1/6 of the whole circumference.

Second one:
The area of the whole circle is (π x the radius²), and
the 40° sector is (40/360) = 1/9 of the whole area.

I hope this is enough help.

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14=h/5<br><br> Help me please

ValentinkaMS [17]

Answer:

14=h/5

h=14*5

h=70

<u>Check</u>:

70/5=14

Hope this helps

6 0

2 years ago

Nine times a number divided by 2 equals 54.

bazaltina [42]

x = the unknown number

$\frac{9x}{2} =54$

nine <u>times</u>(multiplication) a number(x) divided(division) by 2 equals(=) 54,

so 9x is divided by 2

4 0

3 years ago

Find the area. The figure is not drawn to scale.

Ahat [919]

A = 1/2 b h where b = 3.4 cm and h = 8 cm

A = 1/2 (3.4)(8)

A = 13.6 cm^2

Answer: 13.6 cm^2

7 0

3 years ago

Read 2 more answers

Need help on this one too

KiRa [710]

I think it's none because that should be communitive and that's not there so.

4 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