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3 years ago

HELP STAT ITS IN THE PICTURE I HAVE A TIME LIMIT FOR 5 MINS

Mathematics

1 answer:

trapecia [35]3 years ago

4 0

Answer:\

The answers of third one and by the way are you doing an i-Ready Diagnostics?

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Avi, Caleb, Rosa, Sonya, and Zamir compared recent quiz scores and learned that some of their quizzes differed in the number of

never [62]

Answer:

Zamir 75%

Caleb 72%

Rosa 76%

Work

Zamir

.75=75%

75*40=3000

3000/ 100= 30

Zamir got 30 out of 40 questions right

Caleb

30-12=18

Caleb got 18 out of 25 questions right now find percent

18*100=1800

1800/ 25= 72

Caleb got 72%

Rosa

18+1=19

Rosa got 19 out of 25 questions correct now find percent

19*100=1900

1900/ 25=76

Rosa got 76%

5 0

3 years ago

Please help and hurry

lora16 [44]

the correct answer is
16.05

3 0

4 years ago

Read 2 more answers

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$