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3 years ago

What is absolute value?

2 answers:

8 0

Absolute value is when you have either a positive or negative value within a set of "parentheses" or what look just like vertical lines and whatever is within those line stays or becomes positive. The only exception to this is when there is a negative on the outside to act upon the positive values within.

aivan3 [116]3 years ago

5 0

Hello,

Absolute Value: The magnitude of a real number without regard to its sign.

Hope this helps!!!! :)

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12+[ 9 4/5 - ( 1.5 + 2 2/10)]<br><br><br><br>Someone please help me?

lord [1]

Answer:

18.1

<em>The thing You NEEDED to do</em>

<h3><u>Simplify</u> or <u>Evaluate</u> Your Answer</h3>

5 0

3 years ago

Write 95.417 in expanded form with multiplication

Svetllana [295]

In expanded form, it would be 9*10 + 5*1 + 4 * 0.1 + 1 * 0.01 + 7 * 0.001. Hope this helps!

6 0

2 years ago

Can someone explain me how do i do (ii)?

Varvara68 [4.7K]

Answer:

(only ii right?)

Since we have the total number is 360° and our section is 72°, we know that the percentage of students who choose the cinema should be the number that we have divided to the total amount and then multiply the result by 100 ( it's fine if you write 100%, but when you actually calculate it, it's only 100, not 100%):

(72/360) . 100 = 20%

6 0

3 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |<em>subtract $y$ from both sides</em>

$\dfrac{2}{3}x-y+b=0$ |<em>multiply both sides by 3</em>

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |<em>multiply both sides by $\sqrt{13}$ </em>

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |<em>divide both sides by 3</em>

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

2 years ago

20PTS Question On Pic Thx

harina [27]

Answer:

(a) $3800

(b) (a) -- April to May

Step-by-step explanation:

The least amount is found at the lowest point on the graph. That point is in May. It is on the line between 3700 and 3900, so the amount is $3800.

The least donation amount is a month is $3800.

The greatest month-to-month decrease is found where the line on the graph has the steepest negative slope. There are two segments with negative slope:

April - May (decrease of $500)
June - July (decrease of $100)

The decrease from April to May is by far the largest of these two decreases.

The greatest decrease occurred April to May.

6 0

1 year ago

Read 2 more answers