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wel
3 years ago
10

As a salesperson, you are paid $52 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequali

ty for the number of sales you need to make, and describe the solutions.
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

s≥16

Step-by-step explanation:

3s+52≥100

3s≥100-52

3s≥48

s≥48/3

s≥16

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Dmitry_Shevchenko [17]
Chaliye apni maa chudaiye
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3 years ago
Find an equation for the line with the given properties. Express the equation in slope-intercept form.
EleoNora [17]

Answer:

y = 3x - 19

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

Step 1: Define variables

Slope <em>m</em> = 3

Random point (5, -4)

Step 2: Plug in known variables

y = 3x + b

Step 3: Find <em>b</em>

-4 = 3(5) + b

-4 = 15 + b

b = -19

Step 4: Write linear equation

y = 3x - 19

8 0
3 years ago
PLS HELP.... If 4x^2 + bx + 9 = 0, where b is the constant, has exactly one solution, what is a possible value of b ?
IgorLugansk [536]
You can use the discriminant rule, which is :
b^2 -4ac
“exactly one solution” means that it is equal to zero
so:
b^2 -4(4)(9)=0
b^2 -4(36)=0
b^2 -144=0
b^2=144
square root both sides
b=12 or -12

hope this helps !
3 0
3 years ago
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8 0
3 years ago
2logx=3-2log(x+3) solve for x​
AVprozaik [17]

Answer:

\large\boxed{x=\dfrac{-3+\sqrt{40+10\sqrt{10}}}{2}}

Step-by-step explanation:

2\log x=3-2\log(x+3)\\\\Domain:\ x>0\ \wedge\ x+3>0\to x>-3\\\\D:x>0\\============================\\2\log x=3-2\log(x+3)\qquad\text{add}\ 2\log(x+3)\ \text{to both sides}\\\\2\log x+2\log(x+3)=3\qquad\text{divide both sides by 2}\\\\\log x+\log(x+3)=\dfrac{3}{2}\qquad\text{use}\ \log_ab+\log_ac=\log_a(bc)\\\\\log\bigg(x(x+3)\bigg)=\dfrac{3}{2}\qquad\text{use the de}\text{finition of a logarithm}\\\\x(x+3)=10^\frac{3}{2}\qquad\text{use the distributive property}

x^2+3x=10^{1\frac{1}{2}}\\\\x^2+3x=10^{1+\frac{1}{2}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^2+3x=10\cdot10^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\x^2+3x=10\sqrt{10}\qquad\text{subtract}\ 10\sqrt{10}\ \text{from both sides}\\\\x^2+3x-10\sqrt{10}=0\\\\\text{Use the quadratic formula}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\a=1,\ b=3,\ c=-10\sqrt{10}\\\\b^2-4ac=3^2-4(1)(-10\sqrt{10})=9+40\sqrt{10}\\\\x=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2(1)}=\dfrac{-3\pm\sqrt{40+10\sqrt{10}}}{2}\\\\x=\dfrac{-3-\sqrt{10+10\sqrt{10}}}{2}\notin D

6 0
3 years ago
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