# Find the value of the given expression 1/2[4+6(60(-12))]

## Answers

- 2158

Step-by-step explanation:

1/2[4+6(60(-12))] =

= 1/2[4+6(-720)]

= 1/2(4-4320)

= 1/2(-4316)

= - 2158

[tex]1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5[/tex]

[tex]2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1[/tex]

[tex]3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution[/tex]

[tex]4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3[/tex]

[tex]5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7[/tex]

[tex]6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2[/tex]

[tex]7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0[/tex]

[tex]8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0[/tex]

[tex]9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0[/tex]

[tex]10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2[/tex]

[tex]d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0[/tex]

[tex]1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5[/tex]

[tex]2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1[/tex]

[tex]3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution[/tex]

[tex]4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3[/tex]

[tex]5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7[/tex]

[tex]6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2[/tex]

[tex]7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0[/tex]

[tex]8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0[/tex]

[tex]9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0[/tex]

[tex]10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2[/tex]

[tex]d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0[/tex]

7000 (7 thousand)

700 (7 hundred)

20 (2 tens)

2 (2 units)

Step-by-step explanation:

what is the relationship between the values of the given digits?

1. The 7s in 7,700

2. The 2's in 522

From the knowledge of place values;

7,700 could be broken down thus :

7000 + 700 + 0 + 0

The first 7 depicts thousands as it has 3 trailing digits (7000)

The second 7 depicts hundred as it has 2 trailing digits (700)

522 could be broken down thus :

500 + 20 + 2

From 522

The first '2' has one trailing digit = tens

The ending / last digit ia always = Unit value

The relationships in one through three is that they are beside each other in the problem, and, to the naked eye, are the same number.