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Katyanochek1 [597]
4 years ago
10

A rectangular box contains 336 cubic inches. If it is 12 inches long and 7 inches wide, how deep is it?

Mathematics
2 answers:
Marianna [84]4 years ago
7 0
Hello

Your answer is 4 inches deep :))

Hope this helps
laila [671]4 years ago
3 0
4.36in deep rounded to the nearest hundredth<span />
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Which is the most accurate way to estimate 25% of 53?
Alex Ar [27]
You can calculate them out and check which one is the most accurate

so if one of the choices was 1/3 of 50, then you would do:
1/3 in the calculator (should be 0.333...) times 50

And you can compare it with the EXACT number, which is:
0.25*53=13.25

This way you can test all the options
8 0
4 years ago
Perform the division and leave the result in trigonometric form.
Anettt [7]

Answer:

\frac{1}{2} ( cos40° + isin40°)

Step-by-step explanation:

To divide

\frac{r_{1(cosx_{1}+isinx_{1})  } }{r_{2}(cosx_{2}+isinx_{2})   }

= \frac{r_{1} }{r_{2} } [ cos(x₁ - x₂) + isin(x₁ - x₂)

Given

\frac{2(cos90+isin90}{4(cos50+isin50)}

= \frac{2}{4} ( cos(90 - 50)° + isin(90 - 50)°

= \frac{1}{2} ( cos40° + isin40° )

8 0
3 years ago
Right triangles can be proven to be congruent by knowing that the hypotenuse and leg of one are congruent to the hypotenuse and
Slav-nsk [51]
The answer to the question is true
8 0
3 years ago
Read 2 more answers
?/6=1/2 Will get Brainliest
IgorC [24]

Answer:

3/6

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
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