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Alenkinab [10]
3 years ago
11

I need only 5,6,7 i’m so stupid

Mathematics
1 answer:
nordsb [41]3 years ago
3 0
7 is 4.3 which is a terminating decimal
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The glass container is filled to a height of 2.25 inches. What percent of the container is filled with sand?
Makovka662 [10]
The complete question in the attached figure

Part A) How much sand is currently in the container?
[sand  currently in the container]=(5)*(4 1/2)*(2.25)-----> (5)*(4.5)*(2.25)
[sand  currently in the container]=50.625 in³

the answer Part a) is 50.625 in³

Part B) How much more sand could the container hold before?
[sand could the container hold before]=[5*4.5*3]-[50.625]
[sand could the container hold before]=[67.5]-[50.625]------> 16.875 in³

the answer Part B) is 16.875 in³

Part C) What percent of the container is filled with sand?
the volume of container is [5*4.5*3]=67.5 in³
the volume filled with sand=50.625 in³
therefore
if 100%----------------> 67.5 in³
 X---------------------> 50.625 in³
X=(50.625*100)/67.5=75 %

the answer Part C) is 75%





4 0
4 years ago
How many roots (solutions) does the function below have?<br> f(x) = x3 - 6x2 + 3x + 10
STALIN [3.7K]

Answer:

76

Step-by-step explanation:

7 0
3 years ago
What is 75% of 8 hours and 15 minutes
jarptica [38.1K]
Approximately 6 hours and 18 minutes
8 0
3 years ago
Evaluate the function at the specified value of the independent variable and simplify.
AysviL [449]
q(t)=\dfrac{5t^2+6}{t^2}\\\\1.\\q(2)=\dfrac{5\cdot2^2+6}{2^2}=\dfrac{5\cdot4+6}{4}=\dfrac{20+6}{4}=\dfrac{26}{4}=7.5\\\\2.\\q(0)=\dfrac{5\cdot0^2+6}{0^2}=\dfrac{6}{0}-unde fined\\\\3.\\q(-x)=\dfrac{5\cdot(-x)^2+6}{(-x)^2}=\dfrac{5x^2+6}{x^2}
3 0
4 years ago
The product of two consecutive positive odd numbers is 195.By constructing a quadratic equation and solving it, find the two num
Tomtit [17]

Step-by-step explanation:

x = one odd number

the second odd number is (x+2), as both are consecutive.

so,

x × (x + 2) = 195

x² + 2x = 195

x² + 2x - 195 = 0

the formula to solve this is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 2

c = -195

x = (-2 ± sqrt(4 - 4×1×-195))/(2×1) =

= (-2 ± sqrt(4 + 4×195))/2 = (-2 ± sqrt(784))/2 =

= (-2 ± 28)/2 = -1 ± 14

x1 = -1 + 14 = 13

x2 = -1 - 14 = -15 which is invalid as the numbers are positive.

so, the two numbers are 13 and 15.

5 0
3 years ago
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