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2 years ago

6

A car can be rented for $75 per week plus $0.35 per mile. How many miles can be driven if you have at most $110 to spend for wee

kly transport.

1 answer:

sergey [27]2 years ago

8 0

After you subtract the fixed cost of $75 from your budget of $110, you have $35 to spend on miles. At $0.35 per mile, you can afford 100 of them.

You can drive up to 100 miles for $110 or less.

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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:

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

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How would i etimate the first digit of the quotient

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Kate wants to build a fence around her yard. The width of the yard is one-third the length. The perimeter of the yard is 80 feet

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2 years ago

Use factoring to solve the quadratic equation x2 − 2x − 3 = 12.

melomori [17]

Answer:

$x_1=5\\x_2=-3$

Step-by-step explanation:

You have the following quadratic equation given in the problem:

$x^{2}-2x-3=12$

You must make the equation equal to zero, as following:

$x^{2}-2x-3-12=0$

Add like terms:

$x^{2}-2x-15=0$

Now, to factor the equation, you must find two numbers whose sum is -2 and whose product is -15. Therefore, you have:

$(x-5)(x+3)=0\\x_1=5\\x_2=-3$

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