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vagabundo [1.1K]
3 years ago
14

How to solve 3x+5=-13

Mathematics
1 answer:
Serggg [28]3 years ago
6 0
Greetings!

Solve using the properties of equality:
3x+5=-13
Add -5 to both sides.
(3x+5)+(-5)=(-13)+(-5)
3x=-18
Divide both sides by 3.
(3x)/3=(-18)/3
The Answer Is:
x=-6

Hope this helps.
-Benjamin
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What is the slope of this ?
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Answer:

Essentially, slope is rise/run. Rise is how many units up the graph, and run is how far to the left or right it goes. Rise will be negative if it's going down, and run will be negative if it's going left. There's two ways to find the slope in this case.

1. Stick to the more visual aspect of it, and count how many units up from one point to the next point. Point (1,2) is 2 units up from point (0,0). Then you count how many units right or left it goes. In this case, it's 1 towards the right. This means the rise/run comes out to be 2/1, which means the slope=2.

2. Look at the equations. y=mx+b is the linear format for these lines. y is the y coordinate, m is the slope, x is the x coordinate, and b is the y intercept (the point where x=0). These two lines have shared slope values, which means the slope for both of them is 2.

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3 0
3 years ago
Is x = 7 a solution to the equation 3x - 4 = 15? No Yes
Elis [28]

Answer:

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Step-by-step explanation:

8 0
3 years ago
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Tins eats 8 crackers for a snack each day at school.on Friday,she drops 3 and only eats 5. multy​
Pie

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If she only ate 5 she'll need to get 3 more to keep up her streak of eating 8 cracker.

Step-by-step explanation:

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6 0
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1.Arsenic-74 is used to locate brain tumors. It has a half-life of 17.5 days. 90 mg wereused in a procedure. Write an equation t
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1)\text{   }N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) 5.625 mg will be left

Explanation:

1) Half-life = 17.5 days

initial amount of Arsenic-74 = 90 mg

To get the equation, we will use the equation of half-life:

\begin{gathered} N_t\text{ = N}_0(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}} \\ where\text{ N}_t\text{ = amount remaining} \\ N_0\text{ = initial amount} \\ t_{\frac{1}{2}\text{ }}\text{ = half-life} \end{gathered}N_t\text{ = 90\lparen}\frac{1}{2})^{\frac{t}{17.5}}

2) we need to find the remaining amount of Arsenic-74 after 70 days

t = 70

\begin{gathered} N_t=\text{ 90\lparen}\frac{1}{2})^{\frac{70}{17.5}} \\ N_t\text{ = 5.625 mg} \end{gathered}

So after 70 days, 5.625 mg will be left

4 0
1 year ago
What is this Method???
Rina8888 [55]
Standard algorithm i think 
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