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3 years ago

Hi.

Mathematics

1 answer:

NeX [460]3 years ago

5 0

Answer:

(a)

1 sig: 0.005

2 sig: 0.0048

3 sig: 0.00482

(b)

1 sig: 50

2 sig: 50.

3 sig: 50.0

(c)

1 sig: 0.0010

2 sig: 0.00098

3 sig: 0.000981

Step-by-step explanation:

Significant Figures Rules:

Any non-zero digit is significant.
Any trailing zeros after the decimal is significant.
Any zeros between 2 significant digits are significant.
Zeroes before significant numbers in the decimal place are NOT significant; they are placeholders.

(a)

0.004816 - the zeros are placeholders, so they do not count as sig figs.

(b)

50.00168 - the zeros are between 2 significant figures, so they do count as sig figs.

(c)

0.0009812 - the zeros are placeholders, so they do not count as sig figs

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Please help ASAP IM IN DESPREATE NEED OF HELP

Advocard [28]

Answer:

See the attached image for the graph of the first system.

Step-by-step explanation:

Here's how to graph the first system.

Start with the inequality $-y \le -2x-3$ . You can make this easier to work with by multiplying through by -1. Remember to switch the inequality sign when multiplying by a negative number. OK, you get the inequality

$y \ge 2x+3$ .

The graph will be a half-plane -- all the points on one side of a line. The line that is the boundary of the half-plane has an equation: $y=2x+3$ -- just use an = sign instead of the inequality sign.

Graph the line.

The equation of the line is in slope-intercept form: y = mx + b, so you can tell the y-intercept is 3 and the slope is 2 (think of it as a fraction 2/1). Graph the line by going to the point (0, 3) -- the y-intercept -- then use the slope 2/1 interpreted as "rise over run" to go up 2 units and right 1 unit, arriving at the point (1, 5). Draw the line through those points, (0, 3) and (1, 5).

Now you have to decide which side of the line the inequality $y \ge 2x+3$ is describing. To do this, pick a point which is not on the line, plug its coordinates into the inequality; if the result is true, shade the side of the line the point you picked is on (if false, shade the other side!)

An easy point to pick in this case is the origin, (0, 0). Put zeros in for x and y in the inequality, and you'll get the statement $0 \ge 2(0)+3 \, \Rightarrow \, 0 \ge 3$ . That's false, so shade the side of the line not containing the origin. In the image below, the shading is in purple.

All right, now for the other inequality, $x+2\le 0$ . Subtract 2 from both sides and the inequality becomes $x \le -2$ . This, too, graphs as a half-plane whose boundary line has equation $x=-2$ . Graph the line. A line with an equation that has x in it but not y is a vertical line with all its x-coordinates equal to the number on the right side of the equation. This line is vertical and goes through points such as (-2, 0).

Pick a point not on the line (the origin works again). Put the coordinates into the inequality to get $0\le -2$ which is false. Shade the side of the vertical line which does not contain the origin. In the image below, the shading is in black.

Finally, YAY! \o/ , the solutions to the system are all the points in the plane that got shaded twice. In the image, they are the cross-hatched points above the purple line and to left of the black line.

Note: If you get a system with three inequalities, you'll be graphing three half-planes and looking for points that got shaded three times!

Note: One of your questions has the inequalities $x \ge 0$ and $y \ge 0$ in it. These two inequalities say that the x and y coordinates are both positive or zero, confining your attention to Quadrant I in the upper-right part of a graph, above the x-axis and to the right of the y-axis.

7 0

3 years ago

Find the squares of (y+3)

emmasim [6.3K]

Answer:

Since we know that

(a+b)

+2ab+b

(a−b)

−2ab+b

Now,

(i)

(x+3y)

+2×x×3y+(3y)

+6xy+9y

(ii)

(2x−5y)

=(2x)

−2×2x×5y+(5y)

=4x

−20xy+25y

(iii)

(a+

)

+2×a×

)

+1+

(iv)

(2a−1a)=(1a)

(1a)

=1a

Answer

8 0

2 years ago

SOMEONE HELP PLEASE! ASAP!

Luba_88 [7]

When two lines intersect, the angles across from the intersection (the "X") are congruent (equal), and referred to as "vertical angles."
So by the rule of vertical angles are congruent, we can set them equal:
${x}^{2} - 6x = 1\div2 \: x + 42 \\ {x}^{2} - 6x - 1 \div 2 \: x = 42 \\ {x}^{2} - 6.5x = 42$
${x}^{2} - 6.5x - 42 = 0 \\ multiply \: both\: sides \: by \: 2 \: to \: get \\ rid \: of \: the \: decimal \\ 2({x}^{2} - 6.5x - 42) = 2(0)$
$2 {x}^{2} - 13x - 84 = 0$
to factor we need the factors of 84:
1×84, 2×42, 3×28, 4×21, 6×14
Now we need one of those factors ×2 ( from the x^2 term) minus the other factor to equal 13 (the middle term)
it turns out that if we use 4×21: 4×2 = 8
and 21-8 = 13. Woohoo! we got it
So now we have:
$2 {x}^{2} - 13x - 84 = 0 \\ (2x - 21)(x + 4) = 0$
Now set each factor = 0
1) 2x - 21 = 0 and 2) x + 4 = 0
1) 2x - 21 = 0, 2x = 21, x = 21/2 = 10.5
2) x + 4 = 0, x = -4

Now we plug in each answer to check:
1) x = 10.5
${x}^{2} - 6x = {(10.5)}^{2} - 6(10.5) \\ = 110.25 - 63 = 47.25$
1/2x + 42 = 1/2(10.5) + 42 = 5.25 + 42 = 47.25
47.25 = 47.25
BAM SO IT LOOKS LIKE THAT'S IT!!
But let's check 2) first:
2) x = -4
${( - 4)}^{2} - 6( - 4) = 16 + 24 = 40$
1/2x + 42 = 1/2(-4) + 42 = -2 + 42 = 40
40 = 40, so both work!!
But it asks for m<1, and <1 is supplementary because it lies on same line, meaning the sum of both = 189
Therefore m<1 = 180 - 47.25 = 132.75°
OR m<1 = 180 - 40 = 140°

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3 years ago

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My name is Ann [436]

$19705.23 this is the correct answer

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3 years ago

Read 2 more answers

In a circle whose center is O, arc AB contains 100. Find the number of degrees in angle ABO. WILL MARK BRAINLIEST FOR BEST ANSWE

marusya05 [52]

Answer:

It is so easy u should get this right.

First draw a circle and there is an arc that is like forming a triangle inside a circle and then ab angle is 100 degree and outside is the center of the circle. And since u try to find how much is outside it most likely means that 100-60=40

so it b 40.

Step-by-step explanation:

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3 years ago