All categories

3 years ago

Y^2+8y-9= (y+9) (?)

Mathematics

1 answer:

Vika [28.1K]3 years ago

8 0

Answer:

(y+9)(y-1)

Step-by-step explanation:

y^2+9y-8y-9

y(y+9)-1(y+9)

(y+9)(y-1)

You might be interested in

I diver 125. 65 feet below the surface of the ocean and then 16.5 feet further down he then rose 12.45 feet right and solve an e

lisabon 2012 [21]

(-125.65)+(-16.5)+12.45
= -129.7

8 0

4 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

4 years ago

6 points and brainliest answer

kati45 [8]

Well, the angle looks like it is an obtuse angle, right? And we know that obtuse angles are greater than 90 degrees, so wouldn't the answer be A. 115 degree?
I think the answer is 115 degrees. Hope this helps!

4 0

3 years ago

Read 2 more answers

Solve the inequality: 14.3 > 5(z- 2) - 2.1

natta225 [31]

Answer:

z < 5.28

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtract Property of Equality

Step-by-step explanation:

Step 1: Define

14.3 > 5(z - 2) - 2.1

Step 2: Solve for z

Distribute 5: 14.3 > 5z - 10 - 2.1
Subtract: 14.3 > 5z - 12.1
Add 12.1 on both sides: 26.4 > 5z
Divide 5 on both sides: 5.28 > z
Rewrite: z < 5.28

Here we see that any value z smaller than 5.28 would work as a solution to the inequality.

3 0

3 years ago

A rectangular park measuring 32 yards by 24 yards is surrounded by a trail of uniform width. If the area of the park and the tra

cluponka [151]

Answer:

The width = 38 yard

Step-by-step explanation:

Given

Dimension of Park = 32 by 24 yard

Area = 1748 yd²

Required

Find the width of the park

Given that the park is surrounded by a trail;

Let the distance between the park and the trail be represented with y;

Such that, the dimension of the park becomes (32 + y + y) by (24 + y + y) because it is surrounded on all sides

Area of rectangle is calculated as thus;

Area = Length * Width

Substitute 1748 for area; 32 + 2y and 24 + 2y for length and width

The formula becomes

$1748 = (32 + 2y) * (24 +2y)$

Open Bracket

$1748 = 32(24 + 2y) + 2y(24 + 2y)$

$1748 = 768 + 64y + 48y + 4y^2$

$1748 = 768 + 112y + 4y^2$

Subtract 1748 from both sides

$1748 -1748 = 768 -1748 + 112y + 4y^2$

$0 = 768 -1748 + 112y + 4y^2$

$0 = -980 + 112y + 4y^2$

Rearrange

$4y^2 + 112y -980 = 0$

Divide through by 4

$y^2 + 28y - 245 = 0$

Expand

$y^2 + 35y -7y - 245 = 0$

Factorize

$y(y+35) - 7(y + 35) = 0$

$(y-7)(y+35) = 0$

Split the above into two

$y - 7 = 0\ or\ y + 35 = 0$

$y = 7\ or\ y = -35$

But y can't be less than 0;

$So,\ y = 7$

Recall that the dimension of the park is 32 + 2y by 24 + 2y

So, the dimension becomes 32 + 2*7 by 24 + 2*7

Dimension = 32 + 14 yard by 24 + 14 yard

Dimension = 46 yard by 38 yard

Hence, the width = 38 yard

5 0

4 years ago