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2 years ago

I need help with all the answers please and thank you

Mathematics

1 answer:

Lapatulllka [165]2 years ago

7 0

Answer:

I think 7=2, 6=-2, 4=23, 8=-7, 3=9, 1=12, 2=0, 5=7, 10=8, 11=3, 12=-45, 9=25

Step-by-step explanation:

I tried my best

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Tobias sent a chain letter to his friends, asking them to forward the letter to more friends. The number of people who receive t

malfutka [58]

Answer:

$P(t)=37\times (4)^{(t/9.1)}$

Explanation:

The number of emails is multiplied by 4 every 9.1 weeks.

Thus, since Tobias initially sent the chain letter to 37 friends, the number of letters will grow as per this geometric series:

Start: 37 letters

In 9.1 weeks: 37 × 4

In 2 × 9.1 weeks: 37 × (4)²

In 3 × 9.1 weeks: 37 × (4)³

As you see, the exponent is the number of weeks divided by 9.1

Thus, if your variable is the number of weeks, t, then the exponent is t/9.1

And the exponential function, P(t) will be:

$P(t)=37\times (4)^{(t/9.1)}$

3 0

2 years ago

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xenn [34]

Answer:

B.28, 4 + (6*4) = 24 + 4 = 28

6 0

2 years ago

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Which expression is equivalent to 4√24x∧6y÷ 128x∧4y∧5

LenKa [72]

Answer:

Option D. $\sqrt[4]{\frac{3x^{2}}{2y}}$

Step-by-step explanation:

$\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}}$

$\sqrt[4]{(\frac{24}{128})\times (\frac{x^{6}}{x^{4}})\times (\frac{y}{y^{5}})}$

= $\sqrt[4]{(\frac{3}{16})\times {(x)^{6-4}}\times{(y)^{1-5}}}$

= $\sqrt[4]{(\frac{3}{16})\times x^{2}y^{-4}}$

= $\sqrt[4]{\frac{3}{(2)^{4}}\times x\times y^{-4}}$

= $\sqrt[4]{(3\times x^{2)\times (\frac{y^{-1}}{2})^{4}}}$

= $\frac{y^{-1}}{2}\sqrt[4]{3x^{2}}$

= $\sqrt[4]{\frac{3x^{2}}{2y}}$

Option D. $\sqrt[4]{\frac{3x^{2}}{2y}}$ is the correct answer.

5 0

3 years ago

Jamie has 8/10 of a candy bar left over. He wants to split it into 1/3 pieces. How many 1/3 pieces can he make?

Blizzard [7]

The answer is 2. I say that because 3 times 2 is 6, but if you wanted to do 3 thirds, you'd need 9 pieces.

4 0

3 years ago

Please help me. Brainliest xD

masya89 [10]

$\pi \alpha \alpha \beta \\$

3 0

3 years ago

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I need help with all the answers please and thank you ​

I need help with all the answers please and thank you