What is f(–3) for the function f(a) = –2a2 – 5a + 4?

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

9 0

F(−3)=−18+15+4
And the result is:
f(−3)=1
This is the answer.

tino4ka555 [31]3 years ago

4 1

Follow the order of operation (PEMDAS) giving you 31

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If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____

Sauron [17]

<h2>Answer:</h2>

$2\vec{v}-6\vec{u}=(-52,48) \\ \\ ||2\vec{v}-6\vec{u}||=75.28}$

<h2>Step-by-step explanation:</h2>

In this problem we have two vectors:

$\vec{u}=(5,-7) \ and \ \vec{v}=(-11,3)$

So we need to find two things:

$2\vec{v}-6\vec{u}$

and:

$||2\vec{v}-6\vec{u}||$

FIRST:

In this case we have the multiplication of vectors by scalars. A scalar is a simple number, so:

$2\vec{v}-6\vec{u} \\ \\ Replace \ \vec{v} \ and \ \vec{u} \ by \ the \ given \ vectors: \\ \\ 2(-11,3)-6(5,-7) \\ \\ Multiply \ each \ component \ by \ the \ corresponding \ scalar:\\ \\ (2\times (-11),2\times 3)+(-6\times 5,-6\times (-7)) \\ \\ (-22,6)+(-30,42) \\ \\ Sum \ of \ vectors: \\ \\ (-22-30,6+42) \\ \\ \therefore \boxed{(-52,48)}$

SECOND:

If we name:

$\vec{w}=2\vec{v}-6\vec{u}$

Then, $||2\vec{v}-6\vec{u}||$ is the magnitude of the vector $\vec{w}$ . Therefore:

$||\vec{w}||=||2\vec{v}-6\vec{u}|| \\ \\ ||\vec{w}||=||(-52,48)|| \\ \\ ||\vec{w}||=\sqrt{(-58)^2+48^2} \\ \\ ||\vec{w}||=\sqrt{3364+2304} \\ \\ ||\vec{w}||=\sqrt{5668} \\ \\ \boxed{||\vec{w}||=75.28}$