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IRINA_888 [86]
3 years ago
14

there are 9 acorns in each pile. there are 8 piles. how many acorns are there in all? write an equation that can be used to solv

e this problem
Mathematics
2 answers:
Elanso [62]3 years ago
7 0

Answer:

9x8=72 making the answer 72 acorns in total.

statuscvo [17]3 years ago
5 0

Answer:

72 acorns.

Step-by-step explanation:

9 (acorns per pile)

             *

       8 (piles)

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1. Calculate the area of the<br> figure below. Round to the<br> hundredths place when<br> necessary.
a_sh-v [17]

Answer:

The area is 88.39 cm

Step-by-step explanation:

Here, we want to find the area of the semicircle given that the diameter of the semi circle is 15 cm

We start by getting the radius

Mathematically r = D/2

= 15/2 = 7.5

Area of semicircle is area of circle/2

= (pi * r^2)/2

= (22/7 * 7.5^2)/2 = 88.39 cm

5 0
3 years ago
Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is
ohaa [14]

The vectors in S form a basis of P_2 if they are mutually linearly independent and span P_2.

To check for independence, we can compute the Wronskian determinant:

\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0

The determinant is non-zero, so the vectors are indeed independent.

To check if they span P_2, you need to show that any vector in P_2 can be expressed as a linear combination of the vectors in S. We can write an arbitrary vector in P_2 as

p=ax^2+bx+c

Then we need to show that there is always some choice of scalars k_1,k_2,k_3 such that

k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p

This is equivalent to solving

(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c

or the system (in matrix form)

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}

This has a solution if the coefficient matrix on the left is invertible. It is, because

\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}

Then

\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}

\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}

A solution exists for any choice of a,b,c, so the vectors in S indeed span P_2.

The vectors in S are independent and span P_2, so S forms a basis of P_2.

5 0
3 years ago
Pls help will be due soon
gulaghasi [49]

Answer:

radius is 78.5 tell me if I am wrong

3 0
3 years ago
Read 2 more answers
Trinh is catering a wedding that 112 people will attend she is making two cakes a chocolate cake and a carrot cake 5/8 of the gu
erica [24]
70 people will be served carrot cake.
6 0
3 years ago
Read 2 more answers
Pls help its due in a couple of mins
Sergeeva-Olga [200]

Answer:

brainily has the answer

Step-by-step explanation:

3 0
2 years ago
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