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laila [671]
3 years ago
15

PLZ answer my question for God's sake​

Mathematics
1 answer:
timofeeve [1]3 years ago
4 0

Answer:

a. 2x+3z

b. 7xy

c.   \frac{3ab}{2c}

d. 10z-6q

e. Subtract xy from the sum of p and q.

f.  Divide the sum of 3 and y by 5.

g. Subtract 2c from the square root of b.

h. 3x ... 4x ...

i. 3y ... y+5 ... 5y+5

j. length ... 3 ... breadth ... 8b ... 3b²

Step-by-step explanation:

Sum means addition. 2x + 3z

Product means multiplication. 7y × x = 7xy

Divide 3ab by 2c. 3ab ÷ 2c = \frac{3ab}{2c}

Subtract 6q from 10z.  10z-6q

(p+q)-xy : Subtract xy from the sum of p and q.

(3+y)/5: Divide the sum of 3 and y by 5.

\sqrt{b}- 2c: Subtract 2c from the square root of b.

Let the number of boys be x. Girls are three times more so 3 × x = 3x. The total would be 3x + x = 4x.

Let Nora's age be y. Father is three times more so 3 × y = 3y. Brother is 5 more so would be y+5. The total would be y + 3y + y + 5 = 5y+5

3b is three times more than b, the breadth. Thus, the length is three times more than the breadth since it is 3b. The perimeter would be 3b + b + 3b + b = 8b. The area would be 3b × b = 3b²

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the equation is -2x² = 4-3 (x + 1) and the question is justify that it is a 2nd degree equation with the unknown x complete.
Serggg [28]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the equation

-2x² = 4-3 (x + 1)

-2x² = 4-3x-3

-2x² = -3x -7

0 = 2x² -3x -7

We know that the degree of the equation is the highest power of x variable in the given equation.

In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.

Thus, the degree of the equation is 2.

Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.

Let us determine the value 'x'

2x² -3x -7 = 0

Add 7 to both sides

2x^2-3x-7+7=0+7

2x^2-3x=7

Divide both sides by 2

\frac{2x^2-3x}{2}=\frac{7}{2}

x^2-\frac{3x}{2}=\frac{7}{2}

Add (-3/4)² to both sides

x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2

x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}

\left(x-\frac{3}{4}\right)^2=\frac{65}{16}

\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

solving

x-\frac{3}{4}=\sqrt{\frac{65}{16}}

x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}

x-\frac{3}{4}=\frac{\sqrt{65}}{4}

Add 3/4 to both sides

x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}

x=\frac{\sqrt{65}+3}{4}

similarly solving

x-\frac{3}{4}=-\sqrt{\frac{65}{16}}

x=\frac{-\sqrt{65}+3}{4}

So the solution of the equation will have the values of x such as:

x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}

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Answer:

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Step-by-step explanation:

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