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Nikolay [14]
3 years ago
13

How many solutions y=x^2-10x+25

Mathematics
1 answer:
MrRissso [65]3 years ago
7 0
<h2>There are infinite solution of y.</h2>

Step-by-step explanation:

Given,

y = x^2 - 10x + 25

To find, the total number of solutions = ?

∴ y = x^2 - 10x + 25

⇒ y = x^2 - 2(x)(5) + 5^2

⇒ y = (x-5)^{2}

There are infinite solution of y.

Thus, there are infinite solution of y.

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Ethan spends 4 dimes on a plastic crab and then Sophia buys a pink flamingo pen for 1 quarter how much money do they spend in al
slega [8]

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In a sphere, if the radius is tripled, how much more volume do you have?
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3 years ago
You go to an arcade and purchase a card with game credits. After playing 5 games you have 33 credits left. You play 4 more games
Amanda [17]

Answer:

y=-3x+48

Step-by-step explanation:

We will use slope-intercept form of equation to write our equation. The equation of a line in slope-intercept form is: y=mx+b, where m= Slope of the line, b= y-intercept.

To write the equation that represents the number of credits y on the cards after x games, we will find slope of our line.

We have been given that after playing 5 games we have 33 credits left. We play 4 more games and we have 21 credits left. So our points will be (5,33) and (9,21).

Let us substitute coordinates of our both given points in slope formula: m=\frac{y_2-y_1}{x_2-x_1},

m=\frac{21-33}{9-5}

m=\frac{-12}{4}=-3

Now let us substitute m=-3 and coordinates of point (5,33) in slope intercept form of equation to find y-intercept.

33=-3\cdot 5+b    

33=-15+b    

33+15=b    

48=b    

Upon substituting m=-3 and b=48 in slope-intercept form of an equation we will get,

y=-3x+48

Therefore, our desired equation will be y=-3x+48.

6 0
3 years ago
Carlos drew a plan for his garden on a coordinate plane. Rose bushes are located at A(–5, 4), B(3, 4), and C(3, –5)
BartSMP [9]

Given:

A(-5,4)

B(3,4)

C(3,-5)

So point D is:

so point D is (-5,-5)

For AB is

Distance between two point is:

\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}

so distance between A(-5,4) and B(3,4) is:

\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}

So AB is 8 unit apart.

For B(3,4) and C(3,-5).

\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}

So BC is 9 unit apart.

For fourth bush point is (-5,-5) it left of point C(3,-5) is:

\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}

so fourth bush is 8 unit left of C.

For fourth bush(-5,-5) below to point A(-5,4)

\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}

so fourth bush 9 units below of A.

8 0
1 year ago
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