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MariettaO [177]
3 years ago
12

How can estamating products help you to place the decimal correctly

Mathematics
2 answers:
Alexxandr [17]3 years ago
7 0
If we have the problem: 3 times 1.2, and if we estimate the product, we get (3 times 1) 3. If you really do multiply it out, then you get 3.6. But what if you didn't know that, and you put 0.36. 0.36 is say off from our estimated product: 3. Therefore, we know that 0.36 is wrong, and 3.6 is correct (or reasonably correct), because 3.6 is close of 3. I hope this helps! :D
olga55 [171]3 years ago
4 0
It is easier because you can see if your decimal answer is closer to the actual answer
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3 years ago
30 POINTS I WILL GIVE U BRAINLIEST AND THANKS AND 5 STARS please help this is very important this is a big part of my grade
mars1129 [50]

Answer:

1.\:2^x=64, x=\boxed{6},\\2.\:x=\left(\frac{2}{3}\right)^3,x=\boxed{\frac{8}{125}}\\3.\:3\cdot (3^4)=3^x, x=\boxed{5}\\4.\:\frac{16}{25}=x^2,x=\boxed{\frac{4}{5}},

Step-by-step explanation:

1.\\\\2^x=64,\\\log 2^x=\log64,\\x\log 2=\log 64,\\x=\frac{\log 64}{\log 2}=\boxed{6}

2.\\x=\left(\frac{2}{5}\right)^3=\frac{2^3}{5^3}=\boxed{\frac{8}{125}}

3. This problem incorporates an exponent property.

Exponent property used: a^b\cdot a^c=a^{(b+c)}, yielding an answer of \boxed{5}

4.\\\\\frac{16}{25}=x^2,\\x=\sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}}=\boxed{\frac{4}{5}}

3 0
3 years ago
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Y=-4-x; domain = {-6, 2, 7}<br> Find range plz
UNO [17]

Step-by-step explanation:

domain is the valid interval for the x (input) values for the function.

the range is the valid interval for the y (result) values of the function.

so, we have a function definition and a valid domain interval for x. we therefore try the x values and see what result values they create. and that defines the range.

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x = 2

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tan(θ) = \frac{1}{0.6}

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tan(θ) = \frac{h}{25}

⇒ h = 25 · tan(θ)

       = 25 · \frac{1}{0.6}

⇒ h = 41.66... ≈ 41.67

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3 years ago
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