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3 years ago

5

Is 0.75 greater than 0.100?

1 answer:

Naya [18.7K]3 years ago

8 0

Yes because you only need to look at the first decimal for problems like these and the first number you see a seven and a one and seven is greater than one

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Neeeddddddd help :/.......

Angelina_Jolie [31]

Answer:

a

Step-by-step explanation:

7 0

3 years ago

Determine the volume of the figure. Use 3.14 for π. Round your answer to the nearest tenth. Show all of your work.

Ostrovityanka [42]

Answer:

1985 cubic units (written as un. with the cubed exponent)

Step-by-step explanation:

First, you find the volume of the cylinder

Here's the formula:

$V=\pi r^{2}h\\V=volume\\r=radius\\h=height$

Now, you plug in and solve.

$V=(3.14)(7.2^{2})(7.4)\\V=(3.14)(51.84)(7.4)\\V=(162.7776)(7.4)\\V=1204.55424\\rounded:\\V=1204.6$

Then, you find the volume of the dome.

Here's the formula:

$volume/of/a/sphere=\frac{4}{3}\pi r^{3}\\volume/of/a/dome:\\V=\frac{2}{3}\pi r^{3} \\V=volume\\r=radius$

Now, you just plug in and solve.

$V=(\frac{2}{3})(\pi)(r^{3})\\V=(\frac{2}{3})(3.14)(7.2^{3})\\V=(\frac{2}{3})(3.14)(373.248)\\V=(2.093)(373.248)\\V=781.208064\\rounded:\\V=781.2$

Finally, you add them together, and there's your answer!

$1204.6+781.2=1985.8$

Hope this helped!

<em>brainliest?</em>

8 0

3 years ago

A Pitcher of fruit punch holds 2 gallons , If 9 people share the entire pitcher equally how much punch does ecah person get

GuDViN [60]

You would do 2 divided by 9 to get that each person would have 2/9 of a gallon or .22 of a gallon. I hope this helps.

5 0

3 years ago

Help me please I don’t know this

ExtremeBDS [4]

its a m8 a a a a a aa aaa

8 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

3 years ago