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3 years ago

14

. how many 1/2s are in 3

1 answer:

sweet [91]3 years ago

6 0

Answer:

6

Step-by-step explanation:

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BRAINLIEST TO THE PERSON THAT ANSWERS FIRST

Oksana_A [137]

Answer:

A repeating decimal can be written as a fraction using algebraic methods, so any repeating decimal is a rational number.

Marco is wrong cuz its not repeating

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2 years ago

Car rentals involve a $130 flat fee and an additional cost of $31.67 a day. what is the maximum number of days you can rent a ca

Vladimir79 [104]

Represent the number of days by x. With this representation, the variable cost of the rental is 31.67x. The total cost is the sum of the fixed and variable costs. This value should not be more than $500. The equation below shows the relationship.
130 + 31.67x ≤ 500

Solving for x gives x ≤ 11.68
Thus, the maximum number of days to rent the car is only 11 days.

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2 years ago

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I need someone to help me

Harlamova29_29 [7]

the second one 6,6,6 it is a square all the others seem like a triangle

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cfrac%7B3%7D%7B4%7D%20x%20%2B%20%20%5Cfrac%7B5%7D%7B6%7D%20%20%3D%205x%20-%20%20%5Cfra

jonny [76]

Answer:

nos agoi o se

Step-by-step explanation:

6 0

3 years ago

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Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

6 0

3 years ago