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2 years ago

Which graph shows a function where f(2)= 4?

Mathematics

1 answer:

fredd [130]2 years ago

7 0

From the curve given we can see that the graph that shows a function where f(2)= 4 is Graph A.

<h3>Equation of quadratic graph</h3>

A quadratic graph is in form of a parabola opened upward or downward.

According to the question, we are to select a graph that has a coordinate (2, 4) on its curve.

To determine thus, we will locate the value of 2 on the x-axis and value of 4 on the y-intercept and trace whether the intersecting point is on the curve.

From the curve given we can see that the graph that shows a function where f(2)= 4 is Graph A.

Learn more on graph here; brainly.com/question/1523847

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Given that (5,6) is on the graph of f(x), find the corresponding point for the function 4f(x)

Anvisha [2.4K]

Answer:

The corresponding point for the function 4f(x) is (5 , 24)

Step-by-step explanation:

* Lets revise some transformation

- A vertical stretching: is the stretching of the graph away from

the x-axis

- A vertical compression: is the squeezing of the graph toward

the x-axis.

• if k > 1, the graph of y = k • f(x) is the graph of f (x) vertically

stretched by multiplying each of its y-coordinates by k.

• if 0 < k < 1 (a fraction), the graph is f (x) vertically compressed

by multiplying each of its y-coordinates by k.

* In our problem:

- Point (5 , 6) is on the graph of f(x)

- The function will be 4f(x), that means vertical stretching

or vertical compression

∵ k = 4

∵ 4 > 1

∴ It is a vertical stretch

∴ We will multiply the y-coordinate by 4

∴ y' = 4 × 6 = 24

* The corresponding point for the function 4f(x) is (5 , 24)

4 0

3 years ago

Refer to the figure and provide an appropriate name for ACK and FHJ

Fofino [41]

Answer: C

Step-by-step explanation:

They are alternate interior angles.

5 0

2 years ago

PLEASE HELP ME thank you<br><br> answer needs to be in simplest radical form <br><br> find d

sweet [91]

Answer:

$d=10$

Step-by-step explanation:

Pythagorean theorem:

$d^2+d^2=10\sqrt{2}^2,\\2d^2=200,\\d^2=100,\\d=\boxed{10}$

Alternative:

$\cos 45^{\circ}=\frac{d}{10\sqrt{2}},\\\frac{\sqrt{2}}{2}=\frac{d}{10\sqrt{2}},\\2d=10\sqrt{2}\cdot \sqrt{2},\\2d=20,\\d=\boxed{10}$

Alternative (derived from the Pythagorean theorem):

All 45-45-90 triangles have side lengths $x-x-x\sqrt{2}$ , where $x\sqrt{2}$ is the hypotenuse. Since this triangle's hypotenuse is $10\sqrt{2}$ , its sides are $10-10-10\sqrt{2}$ .