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Negative four is multiplied through the quantity x + 8. If 6x is then added to this, the output is 2x + 32. What is the value of x? twice the quantity x + 6 is divided by negative four, the result is 5. Find out the number.
This problem translates to the equation -4 (x + 8) + 6x = 2x + 32. Remember to use parentheses for the expression while the words the quantity are used. Use distributive property on the left side of the equation: -4x - 32 + 6x = 2x + 32. Combine such as terms on the left side of the equation: 2x - 32 = 2x + 32. Subtract 2x from both sides of the equation: 2x - 2x - 32 = 2x - 2x + 32. The two sides are not equal. There is no solution: -32 ≠ 32.
design a synchronous, recycling, MOD-12 counter with D FF''s. Use the states 0000 through 1011 in the counter.
To find out the volume of a cube which measures 3 cm by 3 cm by 3 cm, what formula would you use? The volume of a cube is the length of the side cubed and the length of the sid
probability as that of flipping a coin eight times and getting all the times the same side of the coin.)
L.H.S. =cos 12+cos 60+cos 84 =cos 12+(cos 84+cos 60) =cos 12+2.cos 72 . cos 12 =(1+2sin 18)cos 12 =(1+2.(√5 -1)/4)cos 12 =(1+.(√5 -1)/2)cos 12 =(√5 +1)/2.cos 12 R.H.S =c
a3-a2+a-1
A sample of students had a mean age of 35 years along with a standard deviation of 5 years. A student was randomly picked from a group of 200 students. Determine the probability
Two circles touch externally. The sum of their areas is 58 π cm 2 and the distance between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm) Ans:
Let D(subscript12) = ({x,y : x^2 = e ; y^6 = e ; xy =(y^-1) x}) a) Which of the following subsets are subgroups of D(subscript12) ? Justify your answer. i) {x,y,xy,y^2,y^3,e}
Aaron is installing a ceiling fan in his bedroom. Once the fan is in motion, he requires to know the area the fan will wrap. What formula will he use? The area of a circle is π
If roots of (x-p)(x-q) = c are a and b what will be the roots of (x-a)(x-b) = -c please explain. Solution) (x-p)(x-q)=c x2-(p+q)x-c=0 hence, a+b=p+q and a.b=pq-c
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