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3 years ago

Yesterday, there were 60 problems assigned for math homework. Adele got

Mathematics

2 answers:

Lostsunrise [7]3 years ago

4 0

Answer:

20%

Step-by-step explanation:

12 ÷ 60 = 0.2

0.2 x 100 (to make the percent form) = 20

20%

il63 [147K]3 years ago

3 0

20./. You are welcome

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The Fratellis will be arrested if and only if the police believe Chunk. Question 9 options: 1) p ↔ q 2) p ∨ q 3) p → q 4) ~p 5)

Amiraneli [1.4K]

Answer:

Option 1) p ←→ q

Step-by-step explanation:

The given statement is a combination of two phrases with a connective word.

Phrase 1 - The Fratellis will be arrested.

We consider this phrase as phrase p.

Phrase 2 - If the police believe chunk.

Phrase q.

Now these statements p and q are connected with a connecting word "If only and only".

When two phrases are connected by the connecting words "Only and only if", this connecting word is called as biconditional operator.

p (connecting word) q

p ( If and only if) q

p ←→ q

Therefore, Option 1) p ←→ q is the answer.

3 0

3 years ago

What are the verter and x-intercepts of the graph of the function given below?<br> y=x2-2x-35

horsena [70]

For x intercepts, plug in 0 for y.

0 = (x^2) - 2x - 35

*factoring* = (x-7)(x+5)

x intercepts = 7,-5

As for the vertex, you can use the equation -b/2a for the x-coordinate of the vertex

so,

x = -b/2a = -(-2)/2 = 1

then just find the y value by plugging it back in to the equation.

y = ((1)^2) - 2(1) - 35

= -36

so, vertex is at (1,-36)

5 0

3 years ago

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

3 years ago

Please please help!

Dimas [21]

Answer:

Step-by-step explanation:

Calculate the distance (d) using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 3, 9) and (x₂, y₂ ) = (3, 1)

d = $\sqrt{(3+3)^2+(1-9)^2}$