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3 years ago

(6y+11) (8y+5) what is the value of y

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

3 0

$(6y+11)(8y+5)=0\\ \\ (6y+11)=0\\ \\ 6y=-11\\ \\ y=\frac{-11}{6} \\ \\ (8y+5)=0\\ \\ 8y=-5\\ \\ y=\frac{-5}{8}$

First separate both equations into two separate problems and solve for y.

Final answers:

$y = \frac{-11}{6} and \frac{-5}{8}$

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The perimeter of a playing field for a certain sport is 334ft. The field is a rectangle, and the length is 47 ft longer than the

babunello [35]

Answer:

The dimensions of the rectangle = 60ft by 107ft

Where 60 ft = Width of the playing field

107ft = Length of the playing field

Step-by-step explanation:

A playing field is Rectangular is shape, hence,

The formula for Perimeter of a rectangle = 2(L + W)

P = 334 ft

L = 47 + W

W = W

Hence we input these values in the formula and we have:

334 = 2(47 + W + W)

334 = 2(47 + 2W)

334 = 94 + 4W

334 - 94 = 4W

240 = 4W

W = 240/4

W = 60

There fore, the width of this playing field = 60 ft

The length of this rectangle is calculated as:

47 + W

47 + 60

= 107 ft

The length of this playing field = 107ft

Therefore the dimensions of the rectangle = 60ft by 107ft

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3 years ago

Solve the question and check the solution. Express numbers as integers or simplified fractions. 5(m+5)=7m+9

Yanka [14]

Answer:

m = 8

Step-by-step explanation:

5(m+5)=7m+9 (distribute & evaluate parenthesis)

5(m)+5(5)=7m+9

5m + 25 = 7m + 9 (move all m-terms to one side and constants to the other)

5m - 7m = 9 - 25 (evaluate both sides)

-2m = -16 (divide both sides by -2)

m = (-16) / ( -2) = 8

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4 years ago

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kolbaska11 [484]

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4 years ago

A rectangle has a length of x inches and a width of 10 inches. What is the best equation to represent the perimeter of the recta

Hunter-Best [27]

Perimeter is equal to:

p=2(L+W)

p=2(x+10)

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4 years ago

Read 2 more answers

18. Find the center, vertices, and foci of the ellipse with equation x squared divided by 225 plus y squared divided by 625 = 1.

KengaRu [80]

Answer:

Center (0,0)

Vertices (-15,0), (15,0), (0,-25), (0,25)

Foce (0,-20), (0,20)

Step-by-step explanation:

You are given the ellipse equation

$\dfrac{x^2}{225}+\dfrac{y^2}{625}=1$

The canonical equation of ellipse with center at (0,0) is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

So,

$a^2=225\Rightarrow a=15\\ \\b^2=625\Rightarrow b=25$

Hence, the center of your ellipse is at (0,0) and the vertices are at points (-15,0), (15,0), (0,-25) and (0,25)

This ellipse is strengthen in y-axis, so

$c=\sqrt{b^2-a^2}=\sqrt{625-225}=\sqrt{400}=20$

and the foci are at points (0,-20) and (0,20).

6 0

3 years ago