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Anna35 [415]
2 years ago
5

Solve for the variable: 4+3(x+22)=5x-10

Mathematics
1 answer:
Oduvanchick [21]2 years ago
8 0

When we solve the equation, we will get the real answer <u>x=40</u> as a variable.

<h3>First Degree Equation</h3>

First degree equation is a mathematical sentence, which has values represented by letters.

These letters can indicate a variable or an unknown - which at the end of the equation will be the final value.

— To solve this equation, just: apply the distributive property (multiply the terms that are outside the parentheses, for the terms that are inside the symbol).

— Next, let's add the real numbers together. Next, we'll subtract the like terms before the equality, thus moving the common term after the equality by adding.

— Finally, let's divide the equation, thus obtaining the final result.

4 + 3(x + 22) = 5x - 10

4 + 3x + 66 = 5x - 10

70 + 3x = 5x - 10

3x - 5x = -10 - 70

-2x = -80

x = 80 ÷ 2

<u>x = 40</u>

Therefore, the correct value of X in this equation, will be <u>x = 40</u>.

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3 years ago
If two numbers are in the ratio 5 : 7 and if
abruzzese [7]

Answer:

25 and 35

Step-by-step explanation:

Let the numbers be 5x and 7x

(5x+5)/(7x+5)=3/4

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20-15=21x-20x

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So,the numbers will be

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4 0
3 years ago
What is 8,456 divided by 32
sertanlavr [38]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
The function below has at least one rational zero.
Dmitry_Shevchenko [17]

Answer:

x=-7,x=-1,x=-0.289898,x=0.689898

or x=-7,x=-1,x=\frac{1}{5}-\frac{\sqrt{6}}{5},x=\frac{1}{5}+\frac{\sqrt{6}}{5}

Step-by-step explanation:

<u>Factor by grouping</u>

g(x)=5x^4+38x^3+18x^2-22x-7\\\\g(x)=5x^4+33x^3-15x^2-7x+5x^3+33x^2-15x-7\\\\g(x)=x(5x^3+33x^2-15x-7)+1(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+33x^2-15x-7)\\\\g(x)=(x+1)(5x^3+35x^2-2x^2-14x-x-7)\\\\g(x)=(x+1)(5x^2(x+7)-2x(x+7)-1(x+7))\\\\g(x)=(x+1)(x+7)(5x^2-2x-1)\\\\0=(x+1)(x+7)(5x^2-2x-1)\\

We can easily see that the first two zeroes are x=-1 and x=-7, however, the third zero will need the help of the quadratic formula:

0=5x^2-2x-1\\\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-1)}}{2(5)}\\\\x=\frac{2\pm\sqrt{4+20}}{10}\\ \\x=\frac{2\pm\sqrt{24}}{10}\\ \\x=\frac{2\pm2\sqrt{6}}{10}\\ \\x=\frac{2}{10}\pm\frac{2\sqrt{6}}{10}\\ \\x=\frac{1}{5}\pm\frac{\sqrt{6}}{5}\\\\x_1\approx0.689898\\\\x_2\approx-0.289898

Therefore, the zeroes of the function are x=-7,x=-1,x=-0.289898,x=0.689898

7 0
3 years ago
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