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2 years ago

If (6y − 11)(6y 11) = ay^2 − b, what is the value of a? the value of a is ____________.

1 answer:

7 0

(6y - 11)(6y + 11) = ay² - b |use (a - b)(a + b) = a² - b²

(6y)² - 11² =ay² - b

36y² - 121 = ay² - b |add b to both sides

ay² = 36y² - 121 + b |divide both sides by y² ≠ 0

a = (36y² - 121 + b)/y²

$\boxed{a=\dfrac{36y^2-121+b}{y^2}}$

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If -3 1/4 is subtracted from 1 1/3. What will be the result?

Alenkinab [10]

1_1/3 = 4/3

-3_1/4 = -13/4

(4/3) - (-13/4)

(4/3) + (13/4)

We can see that (4/3) + (13/4) leads to a positive answer.

Answer: choice D

4 0

2 years ago

HELP PLZ DUR 11;59 plz help me

Norma-Jean [14]

Answer:

Base = 10 cm

Height = 60 cm

Step-by-step explanation:

The formula for the area of a triangle is $area=\frac{1}{2} *b*h$ , where <em>b</em> is the length of the base of the triangle, and <em>h</em> is the length of the height of the triangle.

We know the area is 300, and since the height of the triangle is 6 times its base, we know that $h=6*b$ . We can plug in these values into our formula for the area of a triangle, which gives us the following equation to solve:

$300=\frac{1}{2} *b*(6*b)$

$3b^2=300$

$b=\sqrt{100} =10$

The base of the triangle is 10 centimeters.

Now that we know the base of the triangle, we can plug its value in to the original formula to solve for the height of the triangle, which gives us the following equation:

$300=\frac{1}{2} *10*h$

$5h=300$

$h=60$

The height of the triangle is 60 centimeters.

5 0

2 years ago

AABC has vertices at A(1, -9), B(8,0), and C(9,-8).

Rainbow [258]

Check the picture below.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}$

now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.

6 0

2 years ago

Example of an irrational number that is less than 2 and greater than 1.5

disa [49]

For this case, the first thing you should do is take into account the irrational number definition.

An irrational number is one that can not be written as the quotient between two whole numbers.

Their number of decimals is unlimited and they are not periodic.

An irrational number within the specified domain is:

π/2 = 1.570796327

Answer:

an example of an irrational number that is less than 2 and greater than 1.5 is:

π/2 = 1.570796327

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2 years ago

408 rounded to the nearest tenth

andreev551 [17]

408 round to the nearest tenth: 410

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2 years ago