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nataly862011 [7]

3 years ago

13

Akira receives a price at a science fair for having the most informative project her trophy is in the shape of a square pyramid

and is covered in shiny gold foil how much gold foll did it take to cover the trophy including the bottom

1 answer:

mezya [45]3 years ago

7 0

Answer:

216 cm^2

Step-by-step explanation:

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This is Algebra 1 functions and I'm struggling with this one function-

Kitty [74]

Answer:

38

Step-by-step explanation:

f(-9) is the value of f(x) when x = -9. Therefore, f(-9) = 4 from the graph. Doing the same with g(6), we can see that g(6) = 6. Our expression becomes:

-1 * 4 + 7 * 6

= -4 + 42

= 38

7 0

3 years ago

PLEASE HELP ASAP!!!! 15 POINTS! AND I WILL GIVE BRAINLIEST!!!!

pochemuha

Answer:

9

Step-by-step explanation:

7 0

3 years ago

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What type of symmetry does a regular pentagon have

solong [7]

Just a regular pentagon has 5 sides. and also has 5 lines of symmetry.

So if a shape has 7 sides it will have 7 lines of symmetry.

Its always going to be the same number.

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8 0

3 years ago

9 meters of gold wire. How long is the wire in millimeters?

kenny6666 [7]

Milli means 1000
1 meter=1000 milimeters

therefor
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6 0

3 years ago

Read 2 more answers

What is the coefficient of x2y3 in the expansion of (2x + y)5?

Zigmanuir [339]

Option C:

The coefficient of $x^{2} y^{3}$ is 40.

Solution:

Given expression:

$(2 x+y)^{5}$

Using binomial theorem:

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here $a=2 x, b=y$

Substitute in the binomial formula, we get

$(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}$

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}$

$$+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}$

Let us solve the term one by one.

$$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}$

$$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y$

$$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}$

$$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}$

$$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}$

$$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}$

Substitute these into the above expansion.

$(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}$

The coefficient of $x^{2} y^{3}$ is 40.

Option C is the correct answer.

5 0

3 years ago