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Anton [14]
3 years ago
12

Tom and Bob have a total of 49 toys if Bob has 5 more toys than Tom, how many toys does each one have

Mathematics
2 answers:
pickupchik [31]3 years ago
5 0
49 + 5 = 54

So the answer is 54
Andre45 [30]3 years ago
4 0
Let x and y are toys of tom and bob respectively
x+y=49
y=x+5----(bob hs 5 more toys than tom)
 substitute value of y
x+x+5=49
2x+5=49
2x=49-5
2x=44
x=22----tom has toy
y=27------bob has toy
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-27=-3p+15-3p solve the equation and show your work
Gnesinka [82]
-27 = -3p + 15 - 3p

Combine like terms.
-3p -3p = (-3-3)p = -6p

-27 = -6p + 15

Subtract 15 from each side to get the variable term on its own.

-27 - 15 = -6p + 15 - 15

-42 = -6p

Divide each side by -6 to find p.

p = -42 ÷ -6 = 7
6 0
3 years ago
Can a object that measures 1.8 metres by 1.2 metres fit in a gap that measure 1 metre by 0.75 metres
Paha777 [63]
No it cannot because 1.8 and 1.2 are both larger than 0.75 so there is no physical way for it to fit
4 0
3 years ago
PlsAnsweRRRRR<br><br><br><br>The median value is .
Anna11 [10]

Answer:

The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average. ... If there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above.

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Write following ratio in simplest form : <br> 125 g : 1 kg​
Dmitry [639]

Answer:

125g:1Kg

=> 125g : 1000g

<h3>=> 1:4</h3>

5 0
3 years ago
Read 2 more answers
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
3 years ago
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