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3 years ago

Please, please , please help.

Mathematics

2 answers:

Phantasy [73]3 years ago

5 0

Answer:

top 3,6,9

Bottom 60,120 ,180

Step-by-step explanation:

topjm [15]3 years ago

3 0

Answer:

60,120,180

3,6,9

Step-by-step explanation:

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An art class is planning to paint a rectangular mural with an area of 60 square feet. It has to be at least 4 feet high but no m

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The painting could be 5ft high and 12ft long, or it could be 4ft high and 15ft long, or it could also be 6ft high and 10 ft long

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3 years ago

Read 2 more answers

Which are correct representation of the inequality minus 32X -5 less than 52 minus X select two options

Murljashka [212]

Answer:

The value of X is more than equal to -1.83.

Step-by-step explanation:

We need to find the correct representation of the inequality 'minus 32X -5 less than 52 minus X'

LHS of the inequality will be : -32x-5

RHS of the inequality will be : (52-X)

So,

$-32X-5\leq 52-X$

Adding X both sides

$-32X-5+X\leq 52-X+X\\\\-31X-5\leq 52$

Adding 5 to both sides,

$-31X-5+5\leq 52+5\\\\-31X\leq 57$

Dividing both sides by -31.

$X\geq \dfrac{-57}{31}\\\\X\geq -1.83$

So, the value of X is more than equal to -1.83.

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3 years ago

Pick any Martin Luther king speech 2-4 minutes and pick a inspirational reason why you picked that speech

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3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

5 0

3 years ago

Please helpppp i don’t understand

atroni [7]

Answer:

Think the answer will just be 5

Step-by-step explanation:

I don't use step by step so yeah

5 0

3 years ago

Please, please , please help.​

Please, please , please help.