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3 years ago

9

More help plzzz!!!!!!

1 answer:

pantera1 [17]3 years ago

5 0

200 grams, b/c 1.6 kg converted to grams is 1600grams, divided by 8 pots; therefore 200g

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Can yall like--- help????? i suck at geometry

SSSSS [86.1K]

Answer:

you do Length x Width x height

8 3/8 x 6 x 8 1/4 = 414.56

hope this helps :-)

7 0

3 years ago

Kim receives time-and-a-half for any hours worked exceeding the normal 38 hours per week. The normal hourly rate is $13.63. In a

Gala2k [10]

13.63 * 38 = 517.94
671.28 - 517.94 = 153.88
153.88 : 13.63 = 11.25 the number of hours she overworked

3 0

3 years ago

Question: The hypotenuse of a right triangle has a length of 14 units and a side that is 9 units long. Which equation can be use

kramer

Answer:

The hypotenuse is the longest side in a triangle.

a^2=b^2+c^2.

14^2=9^2+c^2.

c^2=196-81.

c^2=115.

c=√115.

c=10.72~11cm

5 0

3 years ago

The bar diagram represents 24 hours in one day and shows a mark at 30%. If you sleep 30% of the day, how many hours do you sleep

Sedbober [7]

30% = 3/10

Number of hours you sleep = 24hours × 3/10
= 24hours ÷ 10 × 3
= 2.4hours × 3
= 7.2hours

Hope this helps! :)

4 0

3 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