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Zigmanuir [339]

2 years ago

6

A parallelogram with sides s and t has perimeter 2s+2t. Kelly is decorating the edge of her parallelogram-shaped craft table. Th

e tabletop has sides that are 30 inches and 14.5 inches.
What is the perimeter of the tabletop?

1 answer:

Vladimir [108]2 years ago

7 0

Answer:

34.5

Step-by-step explanation:

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Is 4/67 grater that 5/777

Anni [7]

Yes because if you divide 4 by 67 you get about 0.0597 and if you divide 5 by 777 then you get about 0.0064 and 0.0597 is greater than 0.0064

6 0

3 years ago

Sedaia [141]

Answer:

$x = 18$

Step-by-step explanation:

Given

$\frac{x}{6} - 7 = -4$

Required

Solve for x

$\frac{x}{6} - 7 = -4$

Add 7 to both sides

$\frac{x}{6} - 7+7 = -4+7$

$\frac{x}{6} = -4+7$

$\frac{x}{6} = 3$

Multiply through by 6

$6 * \frac{x}{6} = 3 * 6$

$x = 3 * 6$

$x = 18$

7 0

2 years ago

30 gallons to 40 gallons

Flura [38]

Answer:

10 or 70

Step-by-step explanation:

10= 40-30=10

70=40+30=70

4 0

2 years ago

Simplify 27^2.<br><br> please (thxx)

adoni [48]

Answer:

729

Step-by-step explanation:

Raise 27 to the power of 2.

729

4 0

2 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

2 years ago