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Lostsunrise [7]
2 years ago
5

Solve the equation. (If all real numbers are solutions, enter REALS. If there is no solution, enter NO SOLUTION.)

Mathematics
2 answers:
lana [24]2 years ago
4 0

Answer:

x = 1

Step-by-step explanation:

\frac { 7 x + 1 } { 16 } = \frac { 1 } { 2 }\\7x+1=\frac{16}{2} \\7x+1=8 \\7x=8-1 \\7x = 7\\x = \frac{7}{7} \\\bigstar \; \boxed{x = 1}

Hope it helps ⚜

abruzzese [7]2 years ago
3 0

Answer: 1

Step-by-step explanation:

16 * 1/2 = 8

7x + 1 = 8

7x = 7

x = 1

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Which expressions are equivalent to <br><br> 2 ln a + 2 ln b - ln a?
ycow [4]

Answer:

Choices 2, 3, and 5  are correct.

Step-by-step explanation:

Complete question is:

Check all that apply.

1. ln ab² - ln a

2. ln a + 2 ln b  

3. ln a² + ln b² - ln a

4. 2 ln ab

5. ln ab²

ANSWER:

The given expression is : 2 ln a + 2 ln b - ln a

It simplifies to: ln a + ln b² = ln ab²

Checking the given options.

1. ln ab² - ln a   = ln (ab²)/a = ln b²

  FALSE

2. ln a + 2 ln b = ln ab²

  TRUE

3. ln a² + ln b² - ln a = ln (a²b²)/a = ln ab²

  TRUE

4. 2 ln ab = ln a²b²

  FALSE

5. ln ab²

  TRUE

4 0
4 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
nevsk [136]

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6 0
2 years ago
Which point on the axis satisfies the inequality
son4ous [18]
<span>y < x means the value of y is less than the value of x the values in your coordinates are (x,y) so plug them in option C is correct 0 < 1</span>
8 0
4 years ago
Find the equation of a line containing the following points. Write the equation in slope-intercept form.
Sindrei [870]

Answer:

y=1/4x + 1.75

Step-by-step explanation:

1-2/2--2 = 1/4

2=1/4+b

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8 0
3 years ago
What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
3 years ago
Read 2 more answers
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