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2 years ago

What is the slope of a line that is perpendicular to the line whose equation is 5y+2x=12?

Mathematics

1 answer:

amid [387]2 years ago

6 0

Answer:

Step-by-step explanation:

slop(m1)=-(cofficient of X/cofficient of y)

them m1 *m2 =-1

then Ans =5/2

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2+2 pls help btw 4 .....

Vsevolod [243]

Answer:

is it 22.

Step-by-step explanation:

is it?

5 0

3 years ago

Read 2 more answers

What is standard form of y=2/5x - 3?

vlabodo [156]

Answer:

Bring 3x to the left: −3x + y = 2. Multiply all by −1: 3x − y = −2. Note: A = 3, B = −1, C = −2. This form: Ax + By + C = 0. is sometimes called "Standard Form", but is more properly called the "General Form".

Step-by-step explanation:

dat is only exsample so i thingk i help u

3 0

3 years ago

Find f'(x) and state the domain of f': f(x) = In (2x^2+1)

-Dominant- [34]

Answer:

$f'(x) = \frac{4x}{2x^2+1}$

Domain: All Real Numbers

General Formulas and Concepts:

Algebra I

Domain is the set of x-values that can be inputted into function f(x)

Calculus

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

f(x) = ln(2x² + 1)

Step 2: Differentiate

Derivative ln(u) [Chain Rule/Basic Power]: $f'(x) = \frac{1}{2x^2+1} \cdot 2 \cdot 2x^{2-1}$
Simplify: $f'(x) = \frac{1}{2x^2+1} \cdot 4x$
Multiply: $f'(x) = \frac{4x}{2x^2+1}$

Step 3: Domain

We know that we would have issues in the denominator when we have a rational expression. However, we can see that the denominator would never equal 0.

Therefore, our domain would be all real numbers.

We can also graph the differential function to analyze the domain.

5 0

3 years ago

What is 2

sergiy2304 [10]

2 3/8 / 1 1/4
= 19/8 / 5/4
=19/8 * 4/5
= 19/10

answer is B 19/10

6 0

3 years ago

Read 2 more answers

The solution set for 6a^2-a-5=0

Alex_Xolod [135]

Answer:

a=1 or a=5/6

Step-by-step explanation:

I'm going to attempt to factor 6a^2-a-5

a=6

b=-1

c=-5

Find two numbers that multiply to be a*c and add to be b.

a*c=-30 =-6(5)

b=-1 =-6+5

So replace -a with -6a+5a in the expression we started with

6a^2-6a+5a-5

now we factor by grouping

6a(a-1)+5(a-1)

(a-1)(6a-5)

Now let's solve the equation:

(a-1)(6a-5)=0

So a=1 or a=5/6

4 0

3 years ago