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2 years ago

X^(2)-2x+7 I need to find Find the vertex and the axis of symmetry.

Mathematics

1 answer:

AleksandrR [38]2 years ago

8 0

Answer:

Vertex: (1, 6)

Axis of symmetry: x = 1

Step-by-step explanation:

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tino4ka555 [31]

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3 years ago

How to answer this question

12345 [234]

Flip the fraction to make the -3 exponet positive then it will be 9x^5/3x^3 which the answer would then be 3x^2

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3 years ago

Read 2 more answers

A ball is dropped from the top of a building that is 250 feet tall. The

Virty [35]

-16t²+250 = 75

-16t² = 75 -250

-16t² = -175

t² = -175 /-16

t² = 175/16

Here there si two solutions : t = $\sqrt{175/16}$ and t = - $\sqrt{175/16}$

However, as time cannot rewind, we will keep only positive solution

So t = $\sqrt{175/16}$

in seconds it will be : 3.3 seconds

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3 years ago

How to solve this problem

rodikova [14]

$\bf \textit{circumference of a circle}\\\\
C=2\pi r\quad 
\begin{cases}
r=radius\\
-----\\
r=10
\end{cases}\implies C=2\pi 10\implies \boxed{C=20\pi }\\\\
-------------------------------\\\\$

$\bf \textit{let's converte 1km to inches}
\\\\\\
km\cdot \cfrac{1000m}{km}\cdot \cfrac{100cm}{m}\cdot \cfrac{in}{2.54cm}\implies \cfrac{1000\cdot 100\cdot in}{2.54}\approx 39370.079\ in\\\\
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\textit{how many times does }20\pi \textit{ go into }39370.079?\implies \boxed{\cfrac{39370.079}{20\pi }}$

3 0

3 years ago

Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p

PIT_PIT [208]

Answer:

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = $x$

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

$\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)$

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

$\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)$

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

$\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8$

So, the answer is:

Colin has 8 sheets left for his third class.

5 0

3 years ago