The perimeter of a rectangular garden is 302 m.

makkiz [27]

$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$

3 0

3 years ago

The solution set for -18 < 5x - 3 is

tensa zangetsu [6.8K]

Answer:

Hope this helps

Step-by-step explanation:

6 0

3 years ago

Let f(x)=3x+5 and g(x)=x^2 find f(x).g(x)

Alina [70]

Assuming you want them multiplied

f(x)*g(x)=(3x+5)(x^2)=3x³+6x²

6 0

3 years ago

XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3

nataly862011 [7]

<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

A = $\frac{1}{2}$ x b x h -------------(i)

Where;

A = 18 $\sqrt{3}$

b = XZ = base of the triangle

h = YZ = height of the triangle = 6

<em>Substitute these values into equation(i) and solve as follows:</em>

18 $\sqrt{3}$ = $\frac{1}{2}$ x b x 6

18 $\sqrt{3}$ = 3b

<em>Divide through by 3</em>

6 $\sqrt{3}$ = b

Therefore, b = XZ = 6 $\sqrt{3}$

<em>Now, assume that the circle is centered at O;</em>

Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180° [sum of angles in a triangle]

=> XOZ + 30° + 30° = 180°

=> XOZ + 60° = 180°

=> XOZ = 180° - 60°

=> XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

$\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}$

$\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}$

$\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}$

$\frac{1}{12} = \frac{1}{2|OZ|}$

$\frac{1}{6} = \frac{1}{|OZ|}$

|OZ| = 6

The radius of the circle is therefore 6.

<em>Now, let's calculate the length of the arc XZ</em>

The length(L) of an arc is given by;

L = θ / 360 x 2 π r ------------------(ii)

Where;

θ = angle subtended by the arc at the center.

r = radius of the circle.

In our case,

θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

5 0

3 years ago

If s(x) = x – 7 and t(x) = 4x2 – x + 3, which expression is equivalent to

Vlada [557]

The answer is If you would like to find the expression that is equivalent to (t*s)(x), you can calculate this using the following steps:

s(x) = x - 7
t(x) = 4x^2 - x + 3

(t*s)(x) = t(s(x)) = t(x - 7) = <span>4(x - 7)^2 - (x - 7) + 3 = 4(x - 7)^2 - x + 7 + 3

The correct result would be </span>4(x – 7)2 – (x – 7) + 3.

6 0

3 years ago