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2 years ago

10

A pizza has a diameter of 14 inches

1 answer:

Natalka [10]2 years ago

6 0

Area = 3.14(r)^2
radius is half of diameter
3.14(7)^2
= 153.86
so B

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What is the answer and work to this problem. <br> 4x+2y=14<br> 2x+y=7

Nimfa-mama [501]

The given system of equation has no solution.

<u>Step-by-step explanation</u>:

<u><em>step 1</em></u><em> :</em>

The given equations are 4x + 2y = 14 and 2x + y = 7.

<u><em>step 2</em></u><em> :</em>

Let 4x + 2y = 14 be the first equation.

Let 2x + y = 7 be the second equation.

The solution (x,y) can be determined by solving the two equations, if only the given two equations are different.

<u><em>step 3</em></u><em> :</em>

In first equation taking 2 out as a common factor on both sides, the equation becomes:

2 (2x + y) = 2 (7)

So, the first equation resembles the second one.

<u><em>step 4</em></u><em> :</em>

Since both the equations are similar, they cannot be solved to get a solution. Therefore, the system of equation has no solution which is also known as inconsistent.

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3 years ago

7. If a = 3 and b = -1, find the value of 2a - 2b.

Ber [7]

Answer:

hola bolilla cómo andas e bueno ay te gachas

5 0

3 years ago

this is due today help please!!!!!! A study involving middle-aged adults was designed to determine the relationship between the

IceJOKER [234]

The answer should be c because the independent variable should not be manipluated.

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3 years ago

What type of line is represented by the equation x=6

Vaselesa [24]

It represented a vertical line that no matter what y is, x is always 6

7 0

3 years ago

Arrange the entries of matrix A in increasing order of their cofactors values

givi [52]

To find the cofactor of

$A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]$

We cross out the Row and columns of the respective entries and find the determinant of the remaining $2\times 2$ matrix with the alternating signs.

$Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|$

$Ac_{11}=4\times 1- -1\times 2$

$Ac_{11}=4+ 2$

$Ac_{11}=6$

$Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|$

$Ac_{12}=-(-7\times 1- -1\times -8)$

$Ac_{12}=-(-7- 8)$

$Ac_{12}=15$

$Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|$

$Ac_{21}=-(5\times 1- 3\times 2)$

$Ac_{21}=-(5-6)$

$Ac_{21}=1$

$A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|$

$Ac_{23}=-(7\times 2 -8\times 5)$

$Ac_{23}=-(14-40)$

$Ac_{23}=26$

$A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|$

$Ac_{31}=5\times -1 -4\times 3$

$Ac_{31}=-5-12$

$Ac_{31}=-17$

$A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|$

$Ac_{33}=7\times 4- -7\times 5$

$Ac_{33}=28+35$

$Ac_{33}=63$

Therefore in increasing order, we have;

$Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63$

7 0

3 years ago

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