3 years ago

GIVEAWAY TO GET 100 POINTS + BRIANLIEST IF YOU ANSWER WHATS 9+10 B)

Mathematics

2 answers:

prohojiy [21]3 years ago

5 0

Answer:

uhhh 21

Step-by-step explanation:

Misha Larkins [42]3 years ago

4 0

Answer:

My guess is 21

Step-by-step explanation:

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Answer:b

Step-by-step explanation:

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3 years ago

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givi [52]

Answer:

Ratio of one dimension:

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3 years ago

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Reil [10]

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3 years ago

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of A is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

4 0

3 years ago

Find the value of x and 3x please!

erica [24]

Answer:

x = 135

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I think this is it :)

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3 years ago