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3 years ago

The width of a rectangle is 1.9 feet. The perimeter is 7.4feet. What is the length?

Mathematics

1 answer:

anygoal [31]3 years ago

5 0

Answer:

7.4=2 (1.9+L)

or,(7.4/2)-1.9=L

Or, 1.8feet=Length

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Can someone help me? After I figure out the sequence i have to write the formula. please help.

sveta [45]

Rewrite the fractions to have a common denominator to find the difference:

2/3 = 4/6

Difference = 4/6-5/6 = -1/6

Sequence formula = an = a1 + d(n-1)

A1 is the first term 5/6

And d = -1/6

Sequence = 5/6 -1/6(n-1)

Simplify to get an =(-n +6)/6

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3 years ago

Angle e measures 126LaTeX: ^\circ∘. What is the measure of LaTeX: \angle∠h?

SVEN [57.7K]

Answer:

$m\angle h=126^o$

Step-by-step explanation:

we know hat

Vertical Angles are the angles opposite each other when two lines cross. They are always congruent.

In this problem

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3 years ago

What is 1/6 + 1/6 + 1/6

tatyana61 [14]

Answer:

1/6 + 1/6 + 1/6 =

1/2 or 0.5

6 0

4 years ago

Read 2 more answers

What is the value of x? 3(x−2)+5x=18 X = ????

ikadub [295]

3(x-2)+5x=18

the correct answer is x=3

first simply the equation

8x=18+6

Now rid of alike terms resulting in

8x=24

now divide 24 by 8

x=3

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3 years ago

Read 2 more answers

SHOW YOUR WORK, PLEASE HELP!!!!!!!!!

Shkiper50 [21]

To find the inverse, we swap the variables y and x, then solve for the new y.

3a. $y=\frac{3}{x-1}$

Swapping the variables: $x=\frac{3}{y-1}$
Solving for y: $x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}$
The domain of this inverse is $x ≠ 0$ .
3b. $y=x^2-1$

Swapping: $x = y^2 - 1$
Solving for y: $y^2 = x + 1 \\ y = \sqrt{x+1}$
The domain of this inverse is $x ≥ -1$ .
3c. $y=\sqrt[3]{\frac{x-7}{3}}$
Swapping: $x=\sqrt[3]{\frac{y-7}{3}}$
Solving for y: $x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7$
The domain of this inverse is all real numbers.
4a. $y=\frac{3}{x-1}$ , $y=1+\frac{3}{x}$
$y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x$
$y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x$

4c. $y=\sqrt[3]{\frac{x-7}{3}}$ , $y=3x^3+7$
$y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x$
$y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x$

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4 years ago