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3 years ago

What is the length of BC? Round to the nearest tenth.

2 answers:

5 0

I think I'm not very sure and I don't want you to get it wrong!!!! i think this 21.3in if you are not sure look for more alternatives

Greeley [361]3 years ago

3 0

Answer:

B) 20.5

Step-by-step explanation:

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Zoe's pizza cost $12.99. She gave a tip of $2.00.

Flura [38]

I think its C , 15% but im not sure

5 0

3 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0

2 years ago

In circle T with m∠STU=78 and ST=8 , find the length of arc SU. Round to the nearest hundredth.

evablogger [386]

Answer:

10.89

Step-by-step explanation:

3 0

2 years ago

The points (-1.6, 2.7) and (3.2, 7.98) are on the graph of a linear relationship between two variables, r and y.

Aleksandr [31]

Answer:

y = 1.1x +4.46

y = 129.86 for x = 114

Step-by-step explanation:

The two-point form of the the equation for a line is useful for this.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (7.98 -2.7)/(3.2 -(-1.6))(x -(-1.6)) + 2.7

y = 5.28/4.8(x +1.6) + 2.7

y = 1.1x +1.76 +2.7

y = 1.1x +4.46

When x=114, the value of y is ...

y = 1.1(114) +4.46

y = 129.86

6 0

2 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

Lostsunrise [7]

Answer:

See below

Step-by-step explanation:

(a) Field lines

A negatively charged particle has an electric field associated with it.

The field lines spread out radially from the centre of the point. They are represented by arrows pointing in the direction that a positive charge would move if it were in the field.

Opposite charges attract, so the field lines point toward the centre of the particle.

For an isolated negative particle, the field lines would look like those in Figure 1 below.

If two negative charges are near each other, as in Figure 2, the field lines still point to the centre of charge.

A positive charge approaching from the left is attracted to both charges, but it moves to the closer particle on the left.

We can make a similar statement about appositive charge approaching from the left.

Thus, there are few field lines in the region between the two particles.

(b) Coulomb's Law

The formula for Coulomb's law is

F = (kq₁q₂)/r²

It shows that the force varies inversely as the square of the distance between the charges.

Thus, the force between the charges decreases rapidly as they move further apart.

5 0

3 years ago