All categories

3 years ago

Which quadratic function has a wider graph than y=2x^2?

Mathematics

1 answer:

Trava [24]3 years ago

8 0

Answer:

y = 1/2 x²

Step-by-step explanation:

The coefficient of the first term in a quadratic, in our case here, x², will tell us how the graph stretches. This is akin to the slope within the linear graph. Similar to the slope, the smaller the coefficient value, or value of slope m, the shallower the angle.

When discussing quadratics, the larger the coefficient of our x² term, the steeper, and skinnier the graph. If we want to look for a graph that is wider than y = 2x², then we need to find a graph with a coefficient that is less than 2.

Our only option then is

y = 1/2 x²

You might be interested in

What are the zeroes of f(x) = x2 − 6x + 8? (4 points) x = −4, 2 x = −4, −2 x = 4, 2 x = 4, −2

babymother [125]

Hey there! :)

Answer:

Third option. x = 2, and x = 4.

Step-by-step explanation:

Find the zeros of this quadratic equation by factoring:

f(x) = x² - 6x + 8

Becomes:

f(x) = (x - 4)(x - 2)

Set each factor equal to 0 to solve for the roots;

x - 4 = 0

x = 4

x - 2 = 0

x = 2

Therefore, the zeros of this equation are at x = 2, and x = 4.

8 0

3 years ago

Read 2 more answers

How do I solve

Maslowich

$Use:\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\\\\sqrt[n]{a}=a^\frac{1}{n}\\------------------------\\\\\left(16n^4\right)^{\frac{5}{4}}=16^\frac{5}{4}\left(n^4\right)^\frac{5}{4}=16^{1\frac{1}{4}}n^{4\cdot\frac{5}{4}}=16^{1+\frac{1}{4}}n^5=16^1\cdot16^\frac{1}{4}\cdot n^5\\\\=16\cdot\sqrt[4]{16}\cdot n^5=16\cdot2\cdot n^5=\boxed{32n^5}\\\\\sqrt[4]{16}=2\ because\ 2^4=16\\\\\text{Answer}\ \boxed{\left(16n^4\right)^\frac{5}{4}=32n^5}$

4 0

3 years ago

The equation c=6.4w represents the cost c for w pounds of walnuts. Does a value of 2.5 for w make sense in this situation? Expla

sineoko [7]

We are given equation: c=6.4w, where the cost c for w pounds of walnuts.

w represents the number of pounds of walnuts.

Number of pounds could be in decimals.

The value of w = 2.5 represents 2.5 pounds of walnuts.

On plugging w=2.5 in c=6.4w equation, we get.

c=6.4×2.5 = $16.

Therefore, for 2.5 pounds of walnuts the total cost is $16.

Hence, a value of 2.5 for w make sense in this situation. It shows that cost of 2.5 pounds of walnuts is $16, where cost of each pound of walnuts is 6.4.

6 0

3 years ago

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 10 N acts on

Step2247 [10]

Answer:

The force is 12 N

Step-by-step explanation:

* Lets explain how to solve the problem

- Direct variation is a relationship between two variables that can

be expressed by an equation in which one variable is equal to a

constant times the other

- If y ∝ x , then y = kx , where k is the constant of variation

* Lets solve the problem

- The force acting on the object varies directly with the object's

acceleration

∵ The force is F in newtons and a is the acceleration is m/s²

∴ F ∝ a

∴ F = ka

- To find k substitute F and a by the initial values of them

∵ A force of 10 N acts on a certain object, the acceleration of the

object is 5 m/s²

∵ F = 10 N when a = 5 m/s²

∵ F = ka

∴ 10 = k(5)

- Divide both sides by 5

∴ K = 2

- Substitute the value of k in the equation

∴ F = 2a

- Lets find the force when the acceleration is 6 m/s²

∵ a = 6 m/s²

∴ F = 2(6) = 12

* The force is 12 N

5 0

3 years ago

Given csc(A) = 60/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational

Katarina [22]

Answer:

$\displaystyle \sec A=\frac{65}{63}$

Step-by-step explanation:

We are given that:

$\displaystyle \csc A=\frac{65}{16}$

Where A is in QI.

And we want to find sec(A).

Recall that cosecant is the ratio of the hypotenuse to the opposite side. So, find the adjacent side using the Pythagorean Theorem:

$a=\sqrt{65^2-16^2}=\sqrt{3969}=63$

So, with respect to A, our adjacent side is 63, our opposite side is 16, and our hypotenuse is 65.

Since A is in QI, all of our trigonometric ratios will be positive.

Secant is the ratio of the hypotenuse to the adjacent. Hence:

$\displaystyle \sec A=\frac{65}{63}$

6 0

3 years ago

Read 2 more answers