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jekas [21]
2 years ago
15

4u^2 - 15u + 18 = 0​

Mathematics
1 answer:
AnnZ [28]2 years ago
6 0

Answer:

Step-by-step explanation:

u=\frac{15 \pm \sqrt{(-15)^2 -4\times 4 \times 18} }{2 \times 4}

u=\frac{15 \pm \sqrt{225-288} }{8} \\u=\frac{15 \pm \sqrt{-63} }{8} \\u=\frac{15 \pm \sqrt{63i^2 } }{8} \\u=\frac{15 \pm 3 \sqrt{7} i }{8}

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Ms. O'Brien makes her own cleaning solution to clean her floors! (yet, she can't clean her measuring cups?) She mixes 8 cups of
Elden [556K]

Answer:

To make a gallons,

Ms. O'Brien needs

12 8/10 cups of water or 12.8 cups of water

3 1/5 cups of water or 3.2 cups of vinegar

Step-by-step explanation:

Ms. O'Brien makes her own cleaning solution to clean her floors! (yet, she can't clean her measuring cups?)

She mixes 8 cups of water with 2 cups of vinegar.

Hence: The total number of cups for the cleaning mixture = 8 cups of water + 2 cups of vinegar = 10 cups

If she wants to make a 1 gallon mixture of her cleaning solution, how much water and how much vinegar should she add?

Note that: There are 16 cups in a gallon

Hence:

For number of cups of water:

10 cups of cleaning mixture = 8 cups of water

16 cups of cleaning mixture = x

Cross Multiply

10x = 16 × 8

x = 16 × 8/10

x = 12 8/10 cups of water or 12.8 cups of water

For the number of cups of vinegar

10 cups of cleaning mixture = 2 cups of water

16 cups of cleaning mixture = x

Cross Multiply

10x = 16 × 2

x = 16 × 2/10

x = 3 1/5 cups of water or 3.2 cups of vinegar

Therefore, to make a gallon

Ms. O'Brien needs

12 8/10 cups of water or 12.8 cups of water

3 1/5 cups of water or 3.2 cups of vinegar

5 0
2 years ago
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
What is 4/8 simplified ​
Vinvika [58]

Answer:

exact form: 1/2

decimal form: 0.5

Step-by-step explanation:

8 0
3 years ago
Which expression is equivalent to |a|<.5
ivolga24 [154]

Answer:

lal lesthan or equal to 5

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
Jenny made five loaves of banana bread that had 1/4 cup of oil in each loaf. After she was done baking, she had 5/8 cup of oil r
-Dominant- [34]
The answer is 1 - 5/8
= 3/8
7 0
3 years ago
Read 2 more answers
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