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mr_godi [17]
3 years ago
8

A rectangular prism has a volume of 32 cm3. If all measurements of the prism are multiplied by 4, what is the new volume?

Mathematics
1 answer:
Likurg_2 [28]3 years ago
5 0

Answer: i think the new volume would be 512 i am not 100% sure

i determine this because 4*3= 16 so 32816=512 i think this i right though if you need to know how i got those numbers i can tell you

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Image attached, please help.
tia_tia [17]

Answer:

 B 3x: 51  = 3x+24: 85

Step-by-step explanation:

Since the triangles are similar

PS             PQ

------- =  -----------

PT                PR


Substitute the values in

3x             3x+24

------- =  -----------

51         85



3x: 51  = 3x+24: 85

4 0
3 years ago
I need help with a math problem
Neko [114]
Which one I can help with anyone

7 0
3 years ago
-
RSB [31]

Answer:

Step-by-step explanation:

3x² - 3x + 7 = 0

a = 3 ; b = -3 and c = 7

D = b² - 4ac = (-3)² - 4*3*7

D = 9 - 84 = - 75

D = 75i²

√D = √75i² = \sqrt{5*5*3 * i * i}= 5i\sqrt{3}

x = \dfrac{-b+\sqrt{D}}{2a}  \ ; \  x=\dfrac{-b-\sqrt{D}}{2a}\\\\ \\x=\dfrac{3+5i\sqrt{3}}{6} ; \ ; x = \dfrac{3-5i\sqrt{3}}{6}\\\\\\x =\dfrac{3}{6}+\dfrac{5i\sqrt{3}}{6} ; \ ; x=\dfrac{3}{6}-\dfrac{5i\sqrt{3}}{6}\\\\\\Solution:\\\\x=\dfrac{1}{2}+\dfrac{5\sqrt{3}}{6}i ; \ ; x=\dfrac{1}{2}-\dfrac{5\sqrt{3}}{6}i

5 0
2 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
4 years ago
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Danny is planting an onion garden. The table shows the size of Danny's garden (in square feet), x, and the number of onions he c
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The unit rate is B) 4
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3 years ago
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