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2 years ago

Write the verbal sentence as an equation. Fourteen minus the product of 3 and a number is 26.

Mathematics

1 answer:

Firdavs [7]2 years ago

8 0

Answer:

14-3x=26

Step-by-step explanation:

When it says "is" and then a number, that means that it is the answer. In this case, 26.

Product means multiplying.

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Someone help me out please

uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

$(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\ (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4$ .

2 step (A): simplify to obtain the final radical term on one side of the equation.

$-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2$ .

3 step (F): raise both sides of the equation to the power of 2 again.

$(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2$ .

4 step (E): simplify to get a quadratic equation.

$4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0$ .

5 step (D): use the quadratic formula to find the values of x.

$D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}$ .

6 step (C): apply the zero product rule.

$x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}$ .

Additional 7 step: check these solutions, substituting into the initial equation.

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3 years ago

Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p

PIT_PIT [208]

Answer:

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = $x$

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

$\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)$

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

$\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)$

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

$\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8$

So, the answer is:

Colin has 8 sheets left for his third class.

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2 years ago