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3 years ago

13

Shannon said she can find the factored form of a trinomial of the form x^2-bx+c from the factors of b. The sum of the factors of

b will equal c. Explain Shannon’s error

1 answer:

AlladinOne [14]3 years ago

6 0

Can you help me no one is helping me

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PLEASE HELPPPP<br> A Kite is a Parallelogram.<br> Always<br> Never<br> Sometimes

ipn [44]

Answer:

Never

Step-by-step explanation:

This statement is false because a kite is a quadrilateral that has two adjacent equal sides (upper side and lower side). In a parallelogram, two opposite sides are equal in measure.

3 0

2 years ago

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Ty has 230 pages of text to read in the next three days. if he averages 25 pages every 1 1/4 hours, how many hours is he going t

sineoko [7]

Answer:

11.5 hours

Step-by-step explanation:

25 pages every 75 minutes (1 1/4)

Thats means 1 page every 3 minutes (divide 25 and 75 by 25 to get this)

230 pages would mean 690 minutes (times the 3 by 230)

690/6 = 11.5

7 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

At a coffee shop, the first 100 customers' orders were as follows. What is the probability that a customer ordered a cold drink

mestny [16]

Answer:

Concept: Probability

You have a dual probability such that they must order a large and then get the probability from there.
So its x/probability of large
If you would have provided the table it would have been clearer

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2 years ago

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Help me please ......

Kruka [31]

Since the block is a cube, the dimentions from corner to corner equal all the same, as long as the pieces are equal, which in this case, they are. :) So basically what you need to do is find the perimeter of one side of the cube, and that's the answer.

8 0

2 years ago