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3 years ago

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Please help me I have tried so hard but nothing is working whoever comments and gives the right answer for each will get brainie

st and 5 stars please help me

1 answer:

Naddika [18.5K]3 years ago

3 0

Answer:

Step-by-step explanation:

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Given: 3x + y = 1.<br> Solve for y. <br> A. y = -3x + 1<br> B.y = -3x - 1<br> C.y = 3x - 1

RUDIKE [14]

Y=-3x+1
So the answer is A. All you're doing is moving the 3x to the opposite side, and because your switching the side, you need to switch the sign. So it's a negative 3x after you solve for y. Solving for y(or any letter) is just getting that letter alone on one side of the equal sign.

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3 years ago

A polygon has a perimeter of 36 inches. Each side of the polygon has exactly 12 inches long. What is the name of the polygon tha

Nikitich [7]

Answer:

Equilateral Triangle

Step-by-step explanation:

Since each side is the same length, divide the perimeter by side length to get the number of sides.

36in ÷ 12in = 3

A triangle has 3 sides.

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3 years ago

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A sequence is defined recursively using the formula f(n + 1) = –0.5 f(n) . If the first term of the sequence is 120, what is f(5

Georgia [21]

$\bf \begin{array}{rrlll} term&\stackrel{f(n+1)=-0.5f(n)}{~\hfill value~ }\\ \cline{1-2} f(1)&120\\ f(2)&-0.5(120)\\ &-60\\ f(3)&-0.5(-60)\\ &30\\ f(4)&-0.5(30)\\ &-15\\ f(5)&-0.5(-15)\\ &7.5 \end{array}$

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3 years ago

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Compra mercancía por 2580000 más el IVA del 19%y retefuente del 3,5% u se cancela en efecyivo

Ainat [17]

Answer:

i dont speak or type spanish

Step-by-step explanation:

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2 years ago

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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;

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

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3 years ago

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