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3 years ago

Which is greater than? .007 1.07

Mathematics

1 answer:

dybincka [34]3 years ago

4 0

Answer:

1.07 > 0.007

Step-by-step explanation:

Greater than

Plz mark as brainliest if correct! Have a nice day!!!

-Lil G

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Someone just answer this math question thingie lol. you’ll get 33 points G

Stels [109]

1) $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Step-by-step explanation:

We need to solve the polynomials.

1) $(-2x+4)+(x-6)$

Solving:

$(-2x+4)+(x-6)\\=-2x+4+x-6\\Combining\,\,like\,\,terms:\\=-2x+x+4-6\\=-x-2$

So, $(-2x+4)+(x-6)$ is simplified into $=-x-2$

2) $(uv-3)+(4uv+1)$

Solving:

$(uv-3)+(4uv+1)$

Expanding:

$=uv-3+4uv+1\\Combining\,\,like\,\,terms:\\=uv+4uv-3+1\\=5uv-2\\$

So, $(uv-3)+(4uv+1)$ is simplified into $5uv-2\\$

3) $(11x-8)-(2x+3)$

Solving:

$(11x-8)-(2x+3)$

Expanding:

$=11x-8-2x-3\\Combining\,\,like\,\,terms:\,\,\\=11x-2x-8-3\\=-9x-11$

So, $(11x-8)-(2x+3)$ is simplified into $-9x-11$

4) $(x^2-4)-(x-2)$

Solving:

$(x^2-4)-(x-2)$

Expanding:

$=x^2-4-x+2\\Combining\,\,like\,\,terms:\\=x^2-x-4+2\\=x^2-x-2$

SO, $(x^2-4)-(x-2)$ is simplified into $x^2-x-2$

5) $(12-3t-7t^2)+(1+3t-t^2)$

Solving:

$(12-3t-7t^2)+(1+3t-t^2)$

Expanding:

$=12-3t-7t^2+1+3t-t^2\\Combining\,\,like\,\,terms:\\=12+1-3t+3t-7t^2-t^2\\=13+0t-8t^2\\=-8t^2+13$

So, $(12-3t-7t^2)+(1+3t-t^2)$ is simplified into $-8t^2+13$

Keywords: Solving Polynomials

Learn more about solving Polynomials at:

#learnwithBrainly

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Tavon has a gift card for $45 that loses $2 for each 30-day period it is not used. He has another gift card for $35 that loses

leva [86]

Answer:

Step-by-step explanation:

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Mice21 [21]

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3 years ago

28 = 7x + 3 - 2x Need all the work shown too solve for X

Salsk061 [2.6K]

Answer:

X = 5

Step-by-step explanation:

28 = 7x + 3 - 2x

28 = 5x + 3

28 - 3 = 5x + 3 - 3

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25/5 = 5x/5

X = 5

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Alexxandr [17]

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