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2 years ago

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

Mathematics

1 answer:

kompoz [17]2 years ago

3 0

Answer:

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tjhStep-by-step explanation:

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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

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) 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