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pishuonlain [190]
3 years ago
8

What are the first five common multiples of 8 and 11?

Mathematics
1 answer:
Lelechka [254]3 years ago
4 0
8x11=88
so it's 88,176,264,352 and 440
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Do you know what you need to do to solve this problem?
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Which expression is equivalent to<br> (2^3)^-5
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Answer:

(2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×-(2×2×2)×(2×2×2)

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Which is the last operation performed when evaluating (8-2x)^2+4 for x=3
Salsk061 [2.6K]
Addition will be the last operation
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Read 2 more answers
0.8) 498 what is the answer​
Lina20 [59]

Step 1: We make the assumption that 498 is 100% since it is our output value.

Step 2: We next represent the value we seek with $x$x​.

Step 3: From step 1, it follows that $100\%=498$100%=498​.

Step 4: In the same vein, $x\%=4$x%=4​.

Step 5: This gives us a pair of simple equations:

$100\%=498(1)$100%=498(1)​.

$x\%=4(2)$x%=4(2)​.

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS

(left hand side) of both equations have the same unit (%); we have

$\frac{100\%}{x\%}=\frac{498}{4}$

100%

x%​=

498

4​​

Step 7: Taking the inverse (or reciprocal) of both sides yields

$\frac{x\%}{100\%}=\frac{4}{498}$

x%

100%​=

4

498​​

$\Rightarrow x=0.8\%$⇒x=0.8%​

Therefore, $4$4​ is $0.8\%$0.8%​ of $498$498​.

7 0
2 years ago
If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo
Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or,          Q = 40 kg × 4200 × 18

or,           Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule    

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0
3 years ago
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