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anzhelika [568]
3 years ago
11

Let p: The integer has one digit.

Mathematics
1 answer:
dezoksy [38]3 years ago
4 0

Answer:

p ↔ q

Step-by-step explanation:

Given :

Let p: The integer has one digit.

Let q: The integer is less than 10.

To Find:Which represents "The integer has one digit if and only if the integer is less than 10”?

Solution:

1. p ∧ q denotes “p and q”

2.p ∨ q denotes “p or q" (or both)

3.p → q denotes “if p then q”

4. Biconditional p ↔ q denotes “p if and only if q"

Thus Biconditional p ↔ q represents "The integer has one digit if and only if the integer is less than 10”

Hence Option D is correct p ↔ q

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25 - 3x = M

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to make a full size cake u use three cups of flour if they wanted to make a cake that was one-half the size how many cups of flo
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I'd say, 1 and 1/2 cup, if you split it in half.
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In Carna's desk drawer, there are 16 paper clips and 20 elastic bands. In Noelia's office supply tray, there are 10 paper clips
Nimfa-mama [501]

Answer:

Carna has a lower ratio of paper clips to elastic bands

0.2

Step-by-step explanation:

step one

given data

In Carna's desk drawer

number of paper clips 16

number of elastic band= 20

ratio of paper clips to elastic bands

16:20= 1/5 =0.2

Noelia's office supply tray

number of paper clips 10

number of elastic band= 13

ratio of paper clips to elastic bands

10:13= 0.76

Carna has a lower ratio of paper clips to elastic bands

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5 0
2 years ago
B) Alex invests $2000 in an account that has a 6% annual rate of growth. To the nearest year, when
wlad13 [49]

Answer:

part 1) 10 years

part 2) 10 years

Step-by-step explanation:

<u><em>The correct question is:</em></u>

Part 1) Alex invests $2000 in an account that has a 6% annual rate of growth compounded annually. To  the nearest year, when will the investment be worth $3600?

Part 2) Alex invests $2000 in an account that has a 6% annual rate of growth compounded continuously. To  the nearest year, when will the investment be worth $3600?

Part 1) we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06\\n=1  

substitute in the formula above

3,600=2,000(1+\frac{0.06}{1})^{t}  

1.8=(1.06)^{t}  

Apply log both sides

log(1.8)=log[(1.06)^{t}]  

Applying property of exponents

log(1.8)=(t)log(1.06)  

t=log(1.8)/log(1.06)  

t=10.09\ years

Round to the nearest year

t=10\ years

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06  

substitute in the formula above

3,600=2,000(e)^{0.06t}  

1.8=[e]^{0.06t}  

Apply ln both sides

ln(1.8)=ln[e]^{0.06t}  

ln(1.8)=(0.06t)ln[e]  

t=ln(1.8)/0.06  

t=9.80\ years

Round to the nearest year

t=10\ years

6 0
2 years ago
Estimate the difference. Round each number to the nearest ten, then subtract.
Nady [450]
110- 120= -10

I think -10 is the answer
5 0
3 years ago
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