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Taya2010 [7]
2 years ago
7

Can someone answer 33 please?!

Mathematics
2 answers:
Temka [501]2 years ago
8 0
(x)+(x+1)+(x+2)=228.

3x+3=228
3x=225
x=75
Integer 1: 75
Integer 2: 76
Integer 3: 77

75+76+77=228

kifflom [539]2 years ago
8 0
The integers will be:
x, x+1, x+2
Add them together:
3x + 3 = 228
Subtract 3 from both sides.
3x = 225
Divide both sides by 3
x = 75
x + 1 = 76
x + 2 = 77
CHECK
75 + 76 + 77 = 228
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Using right triangle below find the sine of angle B
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Answer:

1/2

Step-by-step explanation:

sin = opposite/hypotenuse, so

sin(B) = \frac{\sqrt{7} }{2\sqrt{7} } = 1/2

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3 years ago
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Write an expression for each sentence
Nitella [24]

Answer:

  1. x^2+y
  2. 3(n-7)
  3. 37x-9.85

Step-by-step explanation:

1. If x represents a number, then x^2 represents its square. If y represents a second number, then the sum of that square and the second number is ...

  x^2 + y

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2. The difference of a number (n) and 7 is (n-7). Three times that difference is ...

  3(n - 7)

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3. If x represents a number, then the product of 37 and a number is 37x. 9.85 less than that is ...

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7 0
2 years ago
2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA
professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

A=45e^{-0.0045(t)}

a) To find the initial amount of this substance

At t=0, we get

A=45e^{-0.0045(0)}A=45e^0

We know that e^0=1 ( anything to the power zero is 1)

we get,

A=45

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

\frac{45}{2}=45e^{-0.0045(t)}

\frac{1}{2}=e^{-0.0045(t)}

Taking natural logarithm on both sides we get,

\ln (\frac{1}{2})=-0.0045(t)^{}(-1)\ln (\frac{1}{2})=0.0045(t)\ln (\frac{1}{2})^{-1}=0.0045(t)\ln (2)=0.0045(t)0.6931=0.0045(t)t=\frac{0.6931}{0.0045}t=154.02

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

A=45e^{-0.0045(2500)}A=45e^{-11.25}A=45\times0.000013=0.000585A=0.000585

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0
8 months ago
My question is down there
slavikrds [6]
I don’t really see the question ?
7 0
2 years ago
Four more than seven times a number is 12.” Write an equation that represents the statement. Then solve.
Leona [35]

7x + 4 = 12

7x = 12-4

7x = 8

x = 7/8

3 0
3 years ago
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