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Anna71 [15]
2 years ago
6

HELP PLEASE!! this is 8th grade math btw

Mathematics
1 answer:
natima [27]2 years ago
5 0

Answer:

the 2nd one

Step-by-step explanation:

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___,36 ,108 ,324 ,972
hjlf
12

First figure out of it is a arithmetic or geometric sequence, in this case it is a geometric sequence because you have to multiply, not add to find the next number.

Divide one number, I'll pick 324 by the number before it, 108, you get three.

Now divide 36 by 3, you get 12.

To check, multiply 12 by 3, you get 36, multiply that by 3, you get 108, and so on and so forth.
3 0
2 years ago
Solve for x.<br><br><br>9x+4≤58 or 16x−2&gt;3<br><br><br> x 30<br><br> x 30
Brrunno [24]

Answer:

wow

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
The length of a rectangle is 3 cm less than twice the width. If the perimeter is 42 cm, what is the length of the rectangle?
larisa [96]

Answer:

The length is 13

Step-by-step explanation:

w+w+l+l= perimeter

if the length is 3m less than twice the width then let's do some math.

42 divided 2 =21. so the width and length of one side have to equal 21. For example, let's use 8 as the width. The length would be twice(16) minus three(13) 13 plus 8 is 21. WOW it equals half of it so no we know the length is 13 and the wdith is 8.

6 0
3 years ago
Find the value of x for the right triangle. Round your answer to the nearest hundredth
Anna007 [38]

Answer:38.56

Step-by-step explanation:

tan35=opposite ➗ adjacent

0.7002=27 ➗ x

Cross multiplying we get

0.7002(x)=27

Divide both sides by 0.7002

0.7002x ➗ 0.7002=27 ➗ 0.7002

x=38.56041131

x=38.56 nearest hundredth

7 0
3 years ago
If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo
Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or,          Q = 40 kg × 4200 × 18

or,           Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule    

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0
3 years ago
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