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3 years ago

5

Meghan counted the candies she had in a jar. She discovered that half of the candies were red. If she had 50 red candies, in whi

ch equation does x represent the total number candies in the jar?

1 answer:

Law Incorporation [45]3 years ago

6 0

50x2=x

50 represents the red candies.

2 represents the other half of candies.

x represents the total amount of candies.

Hope this helps!

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Help pls no wrong answers

Westkost [7]

Joey walked 0.5 miles he usually walked 0.7 miles the difference is 0.2 miles

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2 years ago

The bleachers at the football game are 7/8 full, and 1/2 the fans in the bleachers are rooting for the home team. What fraction

ioda

<u>Answer: </u>

The fraction of the bleachers filled with the home team is $\frac{7}{6}$

<u>Solution: </u>

Given that,

The bleachers at the football game are $\frac{7}{8}$

In those bleachers $\frac{1}{2}$ of the fans are rooting for the home team

So, the fans that are filled with the home team is \left(\left(\frac{7}{8}\right) \times\left(\frac{1}{2}\right)\right)

Hence, the required fraction is \left(\left(\frac{7}{8}\right) \times\left(\frac{1}{2}\right)\right)

Removing the brackets we get,

\frac{7 \times 1}{8 \times 2}

= $\frac{7}{6}$

The required fraction is $\frac{7}{6}$

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3 years ago

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:

<u>Given equation</u>:

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

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

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

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2 years ago

write the missing exponent , 1/169 (fraction) = (1/13)_ <-- blank (what exponent goes in?) the picture's there to give an ide

Dafna1 [17]

The answer is 2.

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3 years ago

. Twelve out of 36 students in a class are in the

Tpy6a [65]

36-12=24

The answer is:

12;24 or 1;2 (simplified)

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3 years ago