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slavikrds [6]
3 years ago
14

Someone help please is been so long seems I don’t do this

Mathematics
1 answer:
Verizon [17]3 years ago
4 0

Answer:

0.5 or 1/2

Step-by-step explanation:

Step 1.  40/50−15/50

Step 2.  40-25= 25/50

Step 3.  25/50÷ 5 =5/10

5/10÷ 5 =1/2

4/5−3/10=1/2

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The numbers 32 and 12 are each rounded to the nearest ten and then multiplied to estimate the product. What is the
seropon [69]

Answer:

300

Step-by-step explanation:

if  you round 32 and 12 to the nearest it would be 30 an 10 so u multiply 30 times 10 too get 300

4 0
3 years ago
Write a system of equations (3 points) for the problem below and then solve it The candy shack has 20 lb. Of mixed white and dar
icang [17]

Answer: the mixture contained 15 pounds of white chocolate and 5 pounds of dark chocolate.

Step-by-step explanation:

Let x represent the number of pounds of white chocolate in the candy shack mixture.

Let y represent the number of pounds of dark chocolate in the candy shack mixture.

The candy shack has 20 lb of mixed white and dark chocolate. This means that

x + y = 20

The 20lb mixture is worth $7.50 per pound. This means that the total cost of the mixture is

20 × 7.50 = $150

White chocolate alone sells for $8.00 per pound and dark chocolate sells for 6.00 per pound. This means that

8x + 6y = 150 - - - - - - - - - - - - - -1

Substituting x = 20 - y into equation 1, it becomes

8(20 - y) + 6y = 150

160 - 8y + 6y = 150

- 8y + 6y = 150 - 160

- 2y = - 10

y = - 10/ - 2

y = 5

x = 20 - y = 20 - 5

x = 15

5 0
4 years ago
The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole
lozanna [386]

Answer:

A = 0.75 gram or 1 gram

Step-by-step explanation:

The half-life of carbon 14 is years. How much would be left of an original -gram sample after 2,292 years? (To the nearest whole number).

We can use the following formula for half-life of ^{14}C to find out how much is left from the original sample after 2,292 years:  

A = A_{0}e^{-0.000124t}

where:

<em>A</em> is the amount left of an original gram sample after <em>t</em> years, and  

A_{0} is the amount present at time <em>t</em> = 0.

The half-life of  ^{14}C  is the time <em>t</em> at which the amount present is one-half the amount at time <em>t </em>= 0.

If 1 gram of  ^{14}C is present in a sample,  

 Solve for A when t = 2,292:  

Substituting A_{0}  = 1 gram into the decay equation, and we have:  

A = A_{0}e^{-0.000124t}

A = A_{0}e^{-0.000124(2,292)}

A = 0.75 g or 1 g  

6 0
3 years ago
Please help if you could. <br><br> f/2 + 9 = 4<br> -9 -9<br> f/2 =
dmitriy555 [2]

Answer:

\frac{f}{2} +9=4

Subtract 9 from both sides:

\frac{f}{2}+9-9=4-9

\frac{f}{2} =-5

Multiply both sides by 2:

2*\frac{f}{2} =-5*2

f=-10

8 0
3 years ago
If 3 cot theta= 2, find the value of 2sin theta -3cos theta /2sin theta +3cos theta
sergeinik [125]

Answer:

0

Step-by-step explanation:

Using the trigonometric identity

cotx = \frac{cosx}{sinx}

Given

3cotθ = 2 , then

3\frac{coso}{sino} = 2 ( multiply both sides by sinθ )

3cosθ = 2sinθ ( subtract 3cosθ from both sides )

2sinθ - 3cosθ = 0

Thus

\frac{2sin0-3cos0}{2sin0+3cos0}

= \frac{0}{2sin0+3cos0}

= 0

5 0
3 years ago
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