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yaroslaw [1]
3 years ago
12

Helppppppppppppppppppppppppppppppppppppp

Mathematics
1 answer:
Kryger [21]3 years ago
4 0
The answer to this would be 2
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How do i write a Word phrase to represent the numerical expresion 5+(17-8)
Lerok [7]

Answer:

If i was to read this out loud, i would say: 5 plus the quantity of 17 minus 8.

Step-by-step explanation:

5 + (17-8)

Parenthesis need to be done first! That is why you need to say 'the quantity of...' You also have to add 5 to it all, and you might as well do it before 'the quantity of...' becuz it comes before it in the numerical expression.

YAY! ! ! YOU GOT IT!!! =)

8 0
3 years ago
??????????????????????????????
tamaranim1 [39]

Answer:

your picture doesnt show all the answers but it doesnt matter well just count it

surface area = (length x width) + (width x height) + (height x length) and the answer we times it by 2 (cuz they are 6 sides in a prism)

so 4 x 5 + 5 x 10 + 10 x 4 = 110

110 x 2 = 220in²

8 0
2 years ago
What is 4.7 x 4.75 =
UNO [17]
The answer for this is 22.325
6 0
3 years ago
The function f(x)=Acosh(Cx)+Bsinh(Cx). I need help determining the value of constant C>0 when the second derivative f''(x)=25
Rus_ich [418]

Answer:

C = 5.

Step-by-step explanation:

First, you need to remember that:

For the function:

h(x) = Sinh(k*x)

We have:

h'(x) = k*Cosh(k*x)

and for the Cosh function:

g(x) = Cosh(k*x)

g'(x) = k*Cosh(k*x).

Now let's go to our problem:

We have f(x) = A*cosh(C*x) + B*Sinh(C*x)

We want to find the value of C such that:

f''(x) = 25*f(x)

So let's derive f(x):

f'(x) = A*C*Sinh(C*x) + B*C*Cosh(C*x)

and again:

f''(x) = A*C*C*Cosh(C*x) + B*C*C*Sinh(C*x)

f''(x) = C^2*(A*cosh(C*x) + B*Sinh(C*x)) = C^2*f(x)

And we wanted to get:

f''(x) = 25*f(x) = C^2*f(x)

then:

25 = C^2

√25 = C

And because we know that C > 0, we take the positive solution of the square root, then:

C = 5

6 0
2 years ago
Subtract 3/y-2 - 1/y+3
Helen [10]
To subtract fractions, we need a common denominator.

\frac{3}{y-2} - \frac{1}{y+3}

the common denominator will be (y-2)(y+3)

\frac{3(y+3)-1(y-2)}{(y-2)(y+3)}
\frac{3y + 9 - y + 2}{(y-2)(y+3)}
\frac{2y + 11}{(y-2)(y+3)}

The answer above will be the simplified form.
3 0
3 years ago
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