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Sav [38]
3 years ago
9

Which algebraic expression is equivalent to the expression below? 2(9x+11)+3

Mathematics
2 answers:
timama [110]3 years ago
4 0

Answer:

B: 18x + 25

Step-by-step explanation:

2(9x+11)+3

Distribute the 2

2*9x + 2*11 +3

18x +22 +3

Combine like terms

18x +25

ipn [44]3 years ago
3 0
The answer is

B: 11x +25
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Captain has $30 flame has a $20 note $5 notes . how much more moneny does captain have
SCORPION-xisa [38]

Captain has $30 money

Flame has $20 +$5 =$25

difference =30-25=5

Captain has $5 more money

5 0
3 years ago
What is .00762 as a multiple of a power of 10​
lys-0071 [83]

Step-by-step explanation:

0.00762 = 7.62 \times  {10}^{ - 3}  \\

3 0
3 years ago
The average person's head has about 100,000 strands of hair.
MrMuchimi

Answer:

B: 1 × 10^5

Step-by-step explanation:

B is correct and C is not becasue you have to have the number multiplied by 10 to a power because then you wouldn't know what number to use as the base. I hope this helps. :)

7 0
3 years ago
Read 2 more answers
How many positive integers less than 500 are equal to 5 times an even integer?
arlik [135]

Answer:

490/10=49

Every 10 numbers is one number. I divided by 490 because it says less then 500

Step-by-step explanation:

4 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
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