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3 years ago

ASAP!!! Pls help me out :(

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

7 0

Answer:

B 70.69

Step-by-step explanation:

V=πr^2h

V=π(1.5)^2(10)

which is around 70.68583 where you round it to 70.69

mr_godi [17]3 years ago

3 0

The answer you are looking for is B. 70.69

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Plz plz help is for my brother his in first grade

Mkey [24]

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3 years ago

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1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago

Sarah buys a new car for $26,500 and pays 3% down payment the terms of her loan 4.5% apr over 4 years. How much is her monthly p

Ber [7]

Answer:

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Step-by-step explanation:

4 0

1 year ago

g A program executes a mix of different instruction types. 25% of the instructions require two clock cycles to execute, 20% requ

Tomtit [17]

Answer:

1.8 cycles

Step-by-step explanation:

The average number of clock cycles per instruction is given by the sum of the product of each possible number of cycles by its likelihood.

1 cycle: 50%

2 cycles : 25%

3 cycles : 20%

4 cycles : 5%

$Avg = 0.5*1+0.25*2+0.20*3+0.05*4\\Avg= 1.8\ cycles$

The average number of clock cycles per instruction is 1.8.

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2 years ago