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3 years ago

Which of the following is the equation of a direct variation that has a constant of variation equal to - 1/2?

Mathematics

1 answer:

Sliva [168]3 years ago

3 0

The answer of the question is c

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Find an equivalent expression to this: 3xy + 4x + 8 + ( -2 )

Arlecino [84]

3xy + 4x -16 ...... I hope that helps .

7 0

3 years ago

Can you help me please

igor_vitrenko [27]

Answer:

Step-by-step explanation:

ΔA = 90° - 31°

= 59°

sin A = perpendicular ÷ base

sin 59° = 400 ft ÷ y

1.6643 = 400 ft ÷ y

y = 400 ÷ 1.6643

y = 240.34 ft

<h2>MARK ME AS BRAINLIST</h2>

7 0

3 years ago

Simplify (b^4)^3 A.B^9 B.B^3 C.B D.B^12

natulia [17]

(b^4)^3

To simplify when something is raised to two different exponents, multiply the two exponents together:

4 * 3 = 12

You now have b^12.

The answer is D.

6 0

3 years ago

What is the area of the circle?

AlexFokin [52]

Answer:

530.93

Step-by-step explanation:

area of a circle = πr²

where r = radius

here we are given the diameter

to convert to radius , we divide by 2

so radius = 26 / 2 = 13

so we have A = πr²

==> plug in r = 13

A = π13²

==> evaluate exponent

A = 169π

==> multiply 169 and π

A = 530.93 ( approximately )

4 0

2 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

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

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

6 0

3 years ago