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3 years ago

Help fast pleaseeeee

Mathematics

1 answer:

9966 [12]3 years ago

8 0

we will find the area of triangle with sides 5y,6y and 9y using herons formula

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The ratio of the width of a rectangle to it's length is 3:4. The length of the rectangle is 36 cm. Find the rectangles width.

Fiesta28 [93]

Answer:

27 cm

Step-by-step explanation:

So the ratio you have is

width(x) : length (36)= 3:4

Divide 36 (length) by the ratio length. 36÷4=9

They are multiples of nine, so multiply the width by 9 to get the 3.

3×9=27

You have 27:36

If you want to check it, divide both by 9 and you get 3:4

3 0

3 years ago

First chance you get the best marks

garri49 [273]

Answer:

<h2><em><u>First box and last box.</u></em></h2>

Step-by-step explanation:

<h2><em><u>It is the first box.</u></em></h2>

Distribute 7 to the first number

7 * -3/4 = -21/4

-21/4 = 5 -1/4

Distribute 7 to the second number

7 * -3

= -21

Put the numbers together

<h2><em><u>5 -1/4x -21 is the answer.</u></em></h2>

They separated the parenthesis.

So it is still 7 * -3/4

and

7*-3.

Hope this helps,

Kavitha

7 0

4 years ago

3(2-0.9h)+(-1.3h-4) does anyone know the answer

natta225 [31]

Answer:

-1.3h - 0.7

Step-by-step explanation:

it is an expression not an answer

3 0

3 years ago

Read 2 more answers

Determine whether the data in the table display exponential behavior. Explain why or why not.

disa [49]

Answer:

The Answer is B

Step-by-step explanation:

7 0

3 years ago

I honestly need help with these

Brilliant_brown [7]

9. The curve passes through the point (-1, -3), which means

$-3 = a(-1) + \dfrac b{-1} \implies a + b = 3$

Compute the derivative.

$y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}$

At the given point, the gradient is -7 so that

$-7 = a - \dfrac b{(-1)^2} \implies a-b = -7$

Eliminating $b$ , we find

$(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}$

Solve for $b$ .

$a+b=3 \implies b=3-a \implies \boxed{b = 5}$

10. Compute the derivative.

$y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6$

Solve for $x$ when the gradient is 2.

$x^2 - 5x + 6 = 2$

$x^2 - 5x + 4 = 0$

$(x - 1) (x - 4) = 0$

$\implies x=1 \text{ or } x=4$

Evaluate $y$ at each of these.

$\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}$

$\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}$

11. a. Solve for $x$ where both curves meet.

$\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5$

$\dfrac{x^3}3 - 2x^2 - 9x = 0$

$\dfrac x3 (x^2 - 6x - 27) = 0$

$\dfrac x3 (x - 9) (x + 3) = 0$

$\implies x = 0 \text{ or }x = 9 \text{ or } x = -3$

Evaluate $y$ at each of these.

$A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}$

$B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}$

$C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}$

11. b. Compute the derivative for the curve.

$y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8$

Evaluate the derivative at the $x$ -coordinates of A, B, and C.

$A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}$

$B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}$

$C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}$

12. a. Compute the derivative.

$y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}$

12. b. By completing the square, we have

$12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4$

so that

$\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}$

13. a. Compute the derivative.

$y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}$

13. b. Complete the square.

$3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3$

Then

$\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}$

5 0

2 years ago