$\sf \left[\begin{array}{cc}\sf 4&\sf 6\\ \sf 5 &\sf 8 \\ \sf 3 &\sf -2\end{array}\right]-\left[\begin{array}{cc}\sf 2&\sf 3\\ \sf 1 &\sf 4 \\ \sf -5&\sf3\end{array}\right]$

Just substract corresponding terms

$\\ \sf\longmapsto \left[\begin{array}{cc}\sf 2 &\sf 3\\ \sf 4&\sf4\\ \sf 8&\sf -5\end{array}\right]$

Option B

6 0

3 years ago

Given two different circles O and Q, which of the following cases presents two circles with no (zero) common tangents?

borishaifa [10]

Concentric circles because if a line is tangent to one circle, it would only intersect teh other

3 0

3 years ago

Read 2 more answers

The longer leg of a right triangle is 6cm longer than the shorter leg and also 1cm longer than the hyptenese. find the perimeter

CaHeK987 [17]

X - a leg (x > 6);
x - 6 - a leg;
x + 1 - a <span>hypotenuse;</span>
Use the Pythagorean theorem
$(x+1)^2=x^2+(x-6)^2$

Use: $(a\pm b)^2=a^2\pm2ab+b^2$

$x^2+2x+1=x^2+x^2-12x+36\ \ \ |-x^2-2x-1\\\\x^2-14x+35=0\ \ \ |-35\\\\x^2-14x=-35\\\\x^2-2\cdot x\cdot7=-35\ \ \ |+7^2\\\\x^2-2\cdot x\cdot7+7^2=-35+7^2\\\\(x-7)^2=14\iff x-7=\pm\sqrt{14}\ \ \ |+7\\\\x=7-\sqrt{14} < 6;\ x=7+\sqrt{14}$

$x=7+\sqrt{14}\\x-6=1+\sqrt{14}\\x+1=8+\sqrt{14}$

The perimeter:

$7+\sqrt{14}+1+\sqrt{14}+8+\sqrt{14}=16+3\sqrt{14}$

5 0

3 years ago