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3 years ago

15

Scientists use characteristics to compare stars. Match each characteristic on the left with the statement on the right that uses

it to compare stars.

1 answer:

Llana [10]3 years ago

8 0

Answer:

Explanation:

some are bigger and some are smaller

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What is 94*738^389428394

Lady_Fox [76]

Answer:

undefined

Explanation:

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3 years ago

\o you sell them?” "Fivepence farthing for one—Twopence for two,” the Sheep replied. "Then two are cheaper than one?” Alice said

NemiM [27]

Answer:

unusual and strange

Explanation:

3 0

3 years ago

Fast plz-The mirror check may involve ______________.

barxatty [35]

Answer:

Realigning the mirror

Explanation:

mirrors should be aligned to minimize blind spots, not look at the tires.

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3 years ago

1.0•10^-10 standard form

Drupady [299]

Answer:

$1.0 * 10^{-10} = 0.0000000001$

Explanation:

Given

$1.0 * 10^{-10}$

Required

Convert to standard form

$1.0 * 10^{-10}$

From laws of indices

$a^{-x} = \frac{1}{a^x}$

So, $1.0 * 10^{-10}$ is equivalent to

$1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}$

$1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}$

$1.0 * 10^{-10} = 1.0 * 0.0000000001$

$1.0 * 10^{-10} = 0.0000000001$

Hence, the standard form of $1.0 * 10^{-10}$ is $0.0000000001$

3 0

3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

1)

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

2)

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

3)

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

4)

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

5)

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

6)

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

7)

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

8)

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago