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Masteriza [31]
2 years ago
13

Solve for x: 5x - 2 = 3x + 8

Mathematics
2 answers:
gayaneshka [121]2 years ago
8 0

5x-2=3x+8


X=5

That's your answer
Sliva [168]2 years ago
7 0

Answer:

x = 5

Step-by-step explanation:

To solve for x in 5x - 2 = 3x + 8 start by getting the variables on the same side. To do this, subtract 3x from 5x. (you always do the opposite of the term's operation)

2x - 2 = 8 now add 2 to 8

2x = 10 now divide 2

x = 5

I hope this helped :)

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Allison’s cupcake and cookie budget for her party is $105. She would like to have at least 40 cupcakes and cookies combined, and
FinnZ [79.3K]

Answer:

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

Step-by-step explanation:

We are told she would like to have at least 40 cupcakes and cookies combined.

If a is number of cupcakes and b is number of cookies, then this inequality is;

a + b ≥ 40

Secondly, we are told she would like to have at most 5 more cupcakes than cookies.

Thus, the inequality is;

a + 5 ≤ b

Lastly, we are told that cupcakes sell for $2.50 and cookies for $1.50

Thus, the inequality is;

2.50a + 1.50b ≤ 105

Finally, the 3 inequalities are;

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

4 0
3 years ago
Evaluate the expression below for x = 2, y = -3, and z = -1.
Mariana [72]
B -5
The answer is -5
4 0
3 years ago
A bag of vegetables contains 6 cups, and each serving is 1/2 a cup. How many servings are in a bag
zhannawk [14.2K]

Answer:

12 servings

Step-by-step explanation:

since each serving is 1/2 a cup and if the bag of veggies contains 6 cups, then if you divide 6/0.5 it is 12 servings

6 0
3 years ago
Read 2 more answers
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
3 0
2 years ago
I need help with number 4
Ray Of Light [21]
You will divide the 12 by 28: 12/28 and that equals 3/7
6 0
3 years ago
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