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Arte-miy333 [17]
2 years ago
10

Y= -(x-5)^2 +1 quadratic function

Mathematics
1 answer:
kompoz [17]2 years ago
3 0

Answer:

dnhtdbgfvdcsdxszvfdcxs

tjhStep-by-step explanation:

tbgsrgnyc bshtnyymmmmmtx yfb

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PLEASE ANSWER THIS ASAP
Ksju [112]

Answer:

8,5

Step-by-step explanation:

5*8-2y=30

40-30=2y

10=2y

y=10/2

y=5

6 0
3 years ago
Read 2 more answers
) At a certain college, students are allowed to choose between online and in-person learning. 70% of the students choose online
lubasha [3.4K]

Answer:

Probability = 0.119

Step-by-step explanation:

P (Coronavirus) = P(Person online & corona) or P(Person offline & corona)

(0.70 x 0.02) + (0.30 x 0.35)

0.014 + 0.105

0.119

4 0
2 years ago
Find the length of the chord made by 3x2 - y2 = 3 and y + x – 5 = 0.
MArishka [77]

Answer:

The correct answer is "9√2 units".

Step-by-step explanation:

The given equations are:

⇒ 3x^2-y^2=3...(equation 1)

⇒ y+x-5=0...(equation 2)

According to the 2nd equation, we get

⇒ y+x-5=0

⇒              y=5-x

On substituting the value of "y" in the 1st equation, we get

⇒  3x^2-y^2=3

     3x^2-(5-x)^2=3

     3x^2-(25+x^2-10x)=3

     2x^2+10x-28=0

On taking common, we get

     x^2+5x-14=0

On applying factorization, we get

     x^2+7x-2x-14=0

   x(x+7)-2(x+7)=0

           (x+7)(x-2)=0

                        x+7=0

                              x=-7

or,

                         x-2=0

                                x=2

Now,

The points are:

(-7, 12) = (x₁, y₁)

(2, 3) = (x₂, y₂)

By using the distance formula, the length of chord will be:

= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

On substituting the values in the above formula, we get

= \sqrt{(2-(-7))^2+(3-12)^2}

= \sqrt{(9)^2+(-9)^2}

= \sqrt{81+81}

= 9 \sqrt{2} units

8 0
2 years ago
Somebody please help.
Rina8888 [55]
Hey, what do you need help with ?
8 0
2 years ago
Read 2 more answers
Laws Of Indicies<br>Can someone please help me with this?​
nasty-shy [4]

Answer:

See below

Step-by-step explanation:

1) :  {a}^{5}  {b}^{4}  {c}^{3}  \div  {a}^{2}  {b}^{3} c \\  \\  = {a}^{5 - 2}  {b}^{4 - 3}  {c}^{3 - 1}  \\  \\  = {a}^{3}  {b}^{1}  {c}^{2}  \\  \\ = {a}^{3}  {b}  {c}^{2}  \\  \\  \\2)\:  (x^2 y^3)^3\\\\= x^{2\times 3}y^{3\times 3}\\\|= x^{6}y^{9}\\\\\\3)\:5 {a}^{4}   \div ( {a}^{2}  \times 3a) \\  \\  = 5 {a}^{4}   \div (3 {a}^{2 + 1}  ) \\  \\  = 5 {a}^{4}   \div (3 {a}^{3}  ) \\  \\  =   \frac{5}{3}  {a}^{4 - 3}  \\  \\ =   \frac{5}{3}  {a} \\  \\  \\ 4)\: \frac{14 {a}^{5} }{2 {a}^{3} \times 7 {a}^{4}  }  \\  \\  =   \frac{14 {a}^{5} }{2 \times 7 {a}^{3 + 4} }   \\  \\  =   \frac{ \cancel{14} {a}^{5} }{ \cancel{14}{a}^{3 + 4} }    \\  \\  =   \frac{  {a}^{5} }{ {a}^{7} }   \\  \\  =  {a}^{5 - 7}  \\  \\  =  {a}^{ - 2}  \\  \\  =  \frac{1}{{a}^{2} }  \\  \\

8 0
2 years ago
Read 2 more answers
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