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The image shows a geometric representation of the function f(x) = x^2 + 2x + 3 written in standard form. What is this function w

AlekseyPX

Answer:

f(x) = (x + 1)² + 2 in vertex form ⇒ 3rd answer

Step-by-step explanation:

* Lets revise how to find the vertex form the standard form

- Standard form ⇒ x² + bx + c, where a , b , c are constant

- Vertex form ⇒(x - h)² + k, where h , k are constant and (h , k) is the

vertex point (minimum or maximum) of the function

- At first we must find h and k

- By equating the two forms we can find the value of h and k

* Lets solve the problem

∵ f(x) = x² + 2x + 3 ⇒ standard form

∵ f(x) = (x - h)² + k ⇒ vertex form

- Put them equal each other

∴ x² + 2x + 3 = (x - h)² + k ⇒ open the bracket power 2

∴ x² + 2x + 3 = x² - 2hx + h² + k

- Now compare the like terms in both sides

∵ 2x = -2hx ⇒ cancel x from both sides

∴ 2 = -2h ⇒ divide both sides by -2

∴ -1 = h

∴ The value of h is -1

∵ 3 = h² + k

- Substitute the value of h

∴ 3 = (-1)² + k

∴ 3 = 1 + k ⇒ subtract 1 from both sides

∴ 2 = k

∴ The value of k = 2

- Lets substitute the value of h and k in the vertex form

∴ f(x) = (x - -1)² + 2

∴ f(x) = (x + 1)² + 2

* f(x) = (x + 1)² + 2 in vertex form

5 0

4 years ago

Find the volume of the garden shed

Leno4ka [110]

Answer:

47.3 m³

Step-by-step explanation:

The garden shed is made up of a rectangular prism and a pyramid.

<h3><u>Volume of a rectangular prism</u></h3>

$\begin{aligned}\textsf{Volume of a rectangular prism}&=\sf width \times length \times height\\&=4 \times 4 \times 2\\&=16 \times 2\\&=32\;\; \sf m^3\end{aligned}$

<h3><u>Volume of a pyramid</u></h3>

<u />

From inspection of the given diagram, the slant height of the pyramid is 3.5 m.

Calculate the perpendicular height of the pyramid using Pythagoras Theorem:

$\begin{aligned}\implies a^2+b^2&=c^2\\2^2+h^2&=3.5^2\\h^2&=8.25\\h&=\sqrt{8.25}\;\; \sf m\end{aligned}$

Therefore:

$\begin{aligned}\textsf{Volume of a pyramid}&=\dfrac{\sf length \times width \times height}{3}\\\\&=\dfrac{\sf 4 \times 4 \times \sqrt{8.25}}{3}\\\\& = 15.31883372\;\; \sf m^3\end{aligned}$

<h3><u>Volume of the garden shed</u></h3>

$\begin{aligned}\implies \textsf{Volume of shed}&=\textsf{Volume of rectangular prism}+\textsf{Volume of pyramid}\\&=32+15.31883372\\& = 47.3\;\; \sf m^3\;(nearest\;tenth)\end{aligned}$

6 0

1 year ago

Read 2 more answers

A. Work of 5 Joules is done in stretching a spring from its natural length to 16 cm beyond its natural length. What is the force

kirza4 [7]

Answer:

(a) 62.5 N

(b) 694.4 N/m

(c) 41.7 N

Step-by-step explanation:

Work done in stretching a spring is

$W = \frac{1}{2}Fe=\frac{1}{2}ke^2$ since $F=ke$

F is the applied force, E is the elongation and k is the spring constant.

(a) Here, e = 16 cm = 0.16 m

$W = \frac{1}{2}Fe$

$F = \dfrac{2W}{e} = \dfrac{2\times5}{0.16}=62.5\text{ N}$

(b) Here, e = 12 cm = 0.12 m

$W =\frac{1}{2}ke^2$

$k=\dfrac{2W}{e^2}=\dfrac{2\times5}{0.12^2}=694.4\text{ N/m}$

(c) The restoring force is the same as the applied force.

$F = \dfrac{2W}{e} = \dfrac{2\times5}{0.12}=41.7\text{ N}$

4 0

4 years ago

A 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the

ira [324]

Using Pythagoras theorem, the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.

Let distance from the wall to the foot of the ladder is 'x' feet and the height of the top of the ladder is 'y' feet.

Pythagoras theorem, $x^{2} + y^{2} = (12)^{2}$ --->(1)

Given, $\frac{dx}{dt}= 2feet/second$ at x=3

Put x=3 in Pythagoras theorem equation (1)

$(3)^{2} + y^{2} = 144$

$y^{2} = 144 - 9$

$y^{2}$ = 135

y = 11.61

Derive equation (1) w.r.t to 't'

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ ---->(2)

substitute the value of 'x', 'dx/dt' and 'y' in equation (2), we get the fast of the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

$2(3)(2) + 2 (11.61)\frac{dy}{dt} = 0$