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Ipatiy [6.2K]
3 years ago
7

Multiply what is 109-91

Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
6 0

Answer:

The product of 109 and 91  is 9919.

Step-by-step explanation:

Given two numbers as 109 and 91 .

We have to find the product of two numbers.

To multiply the two numbers lets first write them in simple form so that finding product will be easy.

109 can be written as ( 100+9 ) and 91 can be written as (100 - 9)

Thus, the  product is written as,

109 \times 91=(100+9)\times (100-9)

Using algebraic identity, (a+b)(a-b)=a^2-b^2

\Rightarrow (100+9)\times (100-9)=(100)^2-9^2=10000-81=9919

Thus the product of 109 and 91  is 9919.

xenn [34]3 years ago
3 0
109-91 is 18, but multiplied is 9919
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Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
dexar [7]

Answer:

The value of the constant C is 0.01 .

Step-by-step explanation:

Given:

Suppose X, Y, and Z are random variables with the joint density function,

f(x,y,z) = \left \{ {{Ce^{-(0.5x + 0.2y + 0.1z)}; x,y,z\geq0  } \atop {0}; Otherwise} \right.

The value of constant C can be obtained as:

\int_x( {\int_y( {\int_z {f(x,y,z)} \, dz }) \, dy }) \, dx = 1

\int\limits^\infty_0 ({\int\limits^\infty_0 ({\int\limits^\infty_0 {Ce^{-(0.5x + 0.2y + 0.1z)} } \, dz }) \, dy } )\, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y }(\int\limits^\infty_0 {e^{-0.1z} } \, dz  }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ])  } \, dy  }) \, dx = 1

C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}]  } \, dy  }) \, dx =1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0  }) \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}]   } \, dx = 1

10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}]  } \, dx = 1

50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1

50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1

50C[0+\frac{1}{0.5} ] =1

100C = 1 ⇒ C = \frac{1}{100}

C = 0.01

3 0
2 years ago
What is equivalent to the square root of -27
lions [1.4K]
For the square root of a negative number, you would have to get into imaginary numbers. If you haven't learned about it yet, then the answer would be no real solutions. But if you have, here's how to solve it:

First, identify that 9 x 3 = 27. To after taking the square root, you will get 3√3 (Since the square root of 9 is 3 and there's no square root of 3, so leave it in the √)

However, since it is a negative number, you will have to include the letter i in the answer. i represents the imaginary part of it since you can't have a negative number in the square root.

So your answer will look like 3i√3.
3 0
3 years ago
A 12 ounce mixture of window cleaner is 4% alcohol. How many ounces of the mixture is alcohol
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0.48 ounce is alcohol
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6 0
3 years ago
An account has $26,000 after 15 years. The account received 2.3 percent interest compounded continuously. How much was deposited
Leya [2.2K]

The intital amount deposited was  $18485.82.

<h3>How to find the compound interest?</h3>

If n is the number of times the interested is compounded each year, and 'r' is the rate of compound interest annually, then the final amount after 't' years would be:

a = p(1 + \dfrac{r}{n})^{nt}

An account has $26,000 after 15 years. The account received 2.3 percent interest compounded continuously.

A = 26000

R = 0.02.3

a = p(1 + \dfrac{r}{n})^{nt}

26000 = p(1 + \dfrac{2.3}{100})^{15}\\\\26000 = p \times 1.4064\\\\\P= 18485.82

Therefore, the intital amount deposited was  $18485.82.

Learn more about compound interest here:

brainly.com/question/1329401

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7 0
2 years ago
Help! Can someone please explain!
tankabanditka [31]

Answer:

( 3 x − 2 ) ( x − 2 ) = 0

Set  3 x − 2  equal to  0  and solve for  x .

x = 2 3

Set  x − 2  equal to  0  and solve for  x .

x = 2

The solution is the result of  3 x − 2 = 0  and  x − 2 = 0. x = 2 3 , 2

The result can be shown in multiple forms.

Exact Form: x = 2 3 , 2

Decimal Form: x = 0. ¯ 6, 2

Step-by-step explanation: There you go hope it helps

5 0
3 years ago
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