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3 years ago

10

Least to greatest there's a picture

2 answers:

valina [46]3 years ago

6 0

Least to greatest is going to be your third answer choice! You're welcome

hichkok12 [17]3 years ago

5 0

It would be c because 11/9=1.22, sq5=2.23, sq11=3.32, sq20=4.5, and 2^3=8

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How do I solve the linear equation of 27=6(x)+4(y) over 20=2(x)+5(y)

larisa86 [58]

<u>27 = 6x + 4y </u>= 1 7/20 = 3 4/5<u>
</u>20 = 2x + 5y<u>
</u>

8 0

3 years ago

What is the volume (in cubic units) of a sphere with a radius of 18 units? Assume that π = 3.14 and round your answer to the nea

Nataly [62]

Answer:

The volume of sphere is $24416.64\ cm^3$ .

Step-by-step explanation:

We have,

Radius of a sphere is 18 units.

It is required to find the volume of sphere.

The formula of the volume of sphere in terms of radius is given by :

$V=\dfrac{4}{3}\pi r^3$

Plugging all values in above formula

$V=\dfrac{4}{3}\times 3.14\times (18)^3\\\\V=24416.64\ cm^3$

So, the volume of sphere is $24416.64\ cm^3$ .

7 0

3 years ago

Evaluate the function f(x) = x2 + 3x for both of the following:

marissa [1.9K]

F(x)=x²+3x

f(0)=0²+3*0=0+0=0

Answer: f(0)=0

f(4)=4²+3*4=16+12=28

Answer: f(4)=28

6 0

3 years ago

Given the linear equation 2x + y = 6, perform the necessary operations to put the equation into the proper general form. Explain

likoan [24]

Step-by-step explanation:

The general equation is y = mx + c...

The given equation is 2x + y = 6

Firstly, move everything on the left except the y to the right.i.e.

y = 6 - 2x

Secondly, rearrange the values on the right to look just like the general equation.

y = -2x + 6

The equation is now in proper general form because it matches the format laid down.

3 0

3 years ago

Three consecutive odd integers have a sum of 27. Find the integers.

jek_recluse [69]

$\sf \bf {\boxed {\mathbb {GIVEN:}}}$

Sum of three consecutive odd integers = $27$

$\sf \bf {\boxed {\mathbb {TO\:FIND:}}}$

The values of the three integers.

$\sf \bf {\boxed {\mathbb {SOLUTION:}}}$

$\sf\purple{The\:three\:consecutive \:odd\:integers\:are\:7,\:9\:and\:11.}$

$\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}$

Let us assume the three consecutive odd integers to be $x$ , $(x+2)$ and $(x+4)$ .

As per the condition, we have

$Sum \: \: of \: \: the \: \: three \: \: consecutive \: \: odd \: \: integers = 27$

$➺ \: x + (x + 2) + (x + 4) = 27$

$➺ \: x + x + 2 + x + 4 = 27$

Now, collect the like terms.

$➺ \: (x + x + x) + (2 + 4) = 27$

$➺ \: 3x + 6 = 27$

$➺ \: 3x = 27 - 6$

$➺ \: 3x = 21$

$➺ \: x = \frac{21}{3} \\$

$➺ \: x = 7$

Therefore, the three consecutive odd integers whose sum is $27$ are $\boxed{ 7 }$ , $\boxed{ 9 }$ and $\boxed{ 11 }$ respectively.

$\sf \bf {\boxed {\mathbb {TO\:VERIFY :}}}$

$⇢ 7 + 9 + 11 = 27$

$⇢ 27 = 27$

⇢ L. H. S. = R. H. S.

$\sf\blue{Hence\:verified.}$

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}$

4 0

3 years ago