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4 years ago

Please help!

1 answer:

8 0

The area of the enclosed space is given by the equation
(6+2x)² = 169

We can take the square root of both sides, since 169 is a perfect square:
√(6+2x)² = √169
6 + 2x = 13

Subtract 6 from each side:
6 + 2x - 6 = 13 - 6
2x = 7

Divide both sides by 2:
2x/2 = 7/2
x = 3.5

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Find 62(9) . Express your answer as a fraction in simplest form

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Peter takes an 8-inch by 3-inch sheet of paper and cuts from one corner of the paper to the diagonally opposite corner. now pete

tekilochka [14]

First get the length of the side the was cut (Hypotenuse).

We can use Pythagorean theorem: a^2+b^2 = c^2
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plug into forumla
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3 years ago

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Anni [7]

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Step-by-step explanation:

6 0

3 years ago

-/2 points

Rom4ik [11]

Answer:

Part 1) The domain of the quadratic function is the interval (-∞,∞)

Part 2) The range is the interval (-∞,1]

Step-by-step explanation:

we have

$f(x)=-x^2+14x-48$

This is a quadratic equation (vertical parabola) open downward (the leading coefficient is negative)

step 1

Find the domain

The domain of a function is the set of all possible values of x

The domain of the quadratic function is the interval

(-∞,∞)

All real numbers

step 2

Find the range

The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.

we have a vertical parabola open downward

The vertex is a maximum

Let

(h,k) the vertex of the parabola

The range is the interval

(-∞,k]

Find the vertex

$f(x)=-x^2+14x-48$

Factor -1 the leading coefficient

$f(x)=-(x^2-14x)-48$

Complete the square

$f(x)=-(x^2-14x+49)-48+49$

$f(x)=-(x^2-14x+49)+1$

Rewrite as perfect squares

$f(x)=-(x-7)^2+1$

The vertex is the point (7,1)

therefore

The range is the interval

(-∞,1]

6 0

3 years ago

Need help with finding foci of parabola

mixer [17]

Hello,
Please, see the attached files.
Thanks.

7 0

3 years ago