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lana66690 [7]
3 years ago
11

PLZ HELP!!!!!! (WILL GIVE BRAINLIEST!!!!) Number 4 only

Mathematics
1 answer:
Elan Coil [88]3 years ago
4 0
Figure B, it's volume is 48 while figure A is 36
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Whoever answers first i will mark as brainiest
bija089 [108]

Answer:

B. 7 inches

Step-by-step explanation:

The surface area of a cube is denoted by: A = 6s², where s is the side length.

Here, we know that A = 294 in², so plug this into the equation to solve for s:

A = 6s²

294 = 6s²

s² = 49

s = √49 = 7

The answer is thus B, 7 inches.

7 0
3 years ago
Please help me with this question
alina1380 [7]

Answer:

D

Step-by-step explanation:

See in title it says thousands,

so multiply whatever number in graph by 1000

Rhode Islands shows 45 in graph, so:

45x1000=45,000

5 0
3 years ago
In a pair of similar polygons, corresponding angles are congruent.
Brilliant_brown [7]
This is true based on the theorem corresponding parts of congruent figures are congruent.
3 0
3 years ago
Can someone please help me i need this really bad.
nekit [7.7K]
The answer is the third one
6 0
3 years ago
A candle burned at a steady rate. After 33 minutes, the candle was 11.2 inches tall. Eighteen minutes later, it was 10.75 inches
Amiraneli [1.4K]

Answer:

9.025 inches

Step-by-step explanation:

Let the length be represented by:

L = mt+L_0

Where L is the length of candle at time t

m is the rate at which the candle is burning.

And L_0 is the initial length of the candle.

As per the question statement, let us put the given values in the equation.

11.2 = m\times 33 + L_0 .... (1)\\10.75 = m\times 51+L_0 ..... (2)

Subtracting (1) from (2):

0.45 = -18m\\\Rightarrow m = -0.025

Putting the value of m in the equation (1):

11.2 = -0.025 \times 33+L_0\\\Rightarrow L_0 = 12.025

Therefore, the equation becomes:

L = -0.025t + 12.025

Now, we have to find the height of candle after 2 hours.

2 hours mean 120 minutes.

L = -0.025 \times 120 + 12.025\\\Rightarrow L = 9.025\ inches

Therefore, the height of candle is <em>9.025 inches</em>.

3 0
3 years ago
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