All categories

3 years ago

How would I solve this?

Mathematics

2 answers:

Korvikt [17]3 years ago

5 0

I think its (4a-3)2 squared

Kryger [21]3 years ago

3 0

I would use PEMDAS. You might want to search up what PEMDAS stands for online. I kinda forgot what it stand for.
Hope this helps!

You might be interested in

Someone help me with this I will make you brain

Kazeer [188]

Answer: i believe it is B

Step-by-step explanation:

7 0

3 years ago

2 (h-8)-h = h-16 what is the solution

Svetach [21]

Let's solve your equation step-by-step.2(h−8)−h=h−16Step 1: Simplify both sides of the equation.2(h−8)−h=h−16Simplify: (Show steps)h−16=h−16Step 2: Subtract h from both sides.h−16−h=h−16−h−16=−16
Step 3: Add 16 to both sides.−16+16=−16+160=0Answer:All real numbers are solutions.

6 0

3 years ago

What is negative 3 plus negative 4 minus negative 5

vesna_86 [32]

The answer to your question is -12

6 0

4 years ago

Read 2 more answers

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago

Use the discriminant to describe the roots of each equation. Then select the best description?

Nitella [24]

Answer:

The roots are real and irrational

Step-by-step explanation:

* Lets explain what is the discriminant

- In the quadratic equation ax² + bx + c = 0, the roots of the

equation has three cases:

1- Two different real roots

2- One real root or two equal real roots

3- No real roots means imaginary roots

- All of these cases depend on the value of a , b , c

∵ The rule of the finding the roots is

x = [-b ± √(b² - 4ac)]/2a

- The effective term is √(b² - 4ac) to tell us what is the types of the root

# If the value under the root b² - 4ac positive means greater than 0

∴ There are two different real roots

# If the value under the root b² - 4ac = 0

∴ There are two equal real roots means one real root

# If the value under the root b² - 4ac negative means smaller than 0

∴ There is real roots but the roots will be imaginary roots

∴ We use the discriminant to describe the roots

* Lets use it to check the roots of our problem

∵ x² - 5x - 4 = 0

∴ a = 1 , b = -5 , c = -4

∵ Δ = b² - 4ac

∴ Δ = (-5)² - 4(1)(-4) = 25 + 16 = 41

∵ 41 > 0

∴ The roots of the equation are two different real roots

∵ √41 is irrational number

∴ The roots are real and irrational

* Lets check that by solving the equation

∵ x = [-(-5) ± √41]/2(1) = [5 ± √41]/2

∴ x = [5+√41]/2 , x = [5-√41]/2 ⇒ both real and irrational

5 0

3 years ago

Read 2 more answers