All categories

3 years ago

Please please please help me out

Mathematics

2 answers:

Murrr4er [49]3 years ago

8 0

Answer:

what grade you in do you even do your work this easy is b

Anna [14]3 years ago

6 0

Answer:

Step-by-step explanation:

The order of operations is also referred to as PEMDAS

P for parenthesis

E for exponents

M for multiplication

D for division

A for addition

S for subtraction

The correct order of operations is a

You might be interested in

A mistake was made in the steps shown to simplify the expression. Which step includes the mistake?

kolezko [41]

Step 2 how did the 8 get there?

6 0

3 years ago

I need a tape diagram that goes with Sadie is saving all her money to buy a horse which will cost her $600. She has already save

ivolga24 [154]

Answer:

407

Step-by-step explanation:

8 0

3 years ago

What is the approximate distance between the points (1,-2) and (-9,3) on a coordinate grid

jeka57 [31]

Answer

$5\sqrt{5}$ ≈ 11.1803398874989485

Detailed Explanation

We will be using the distance formula for this problem

$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Now, let's input the numbers into the formula.

$\sqrt{\left(-9-1\right)^2+\left(3-\left(-2\right)\right)^2}$

Solve.

Therefore, the distance between the two points is

$5\sqrt{5}$ ≈ 11.1803398874989485

3 0

3 years ago

Read 2 more answers

Find the point on the line y=2x that is closest to the point (4, 5) .

alexgriva [62]

That point would be : (2.5,5)

7 0

3 years ago

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

leonid [27]

Answer-

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

$12x-5y+14=0 \\ 12x-5y-14=0$

Solution-

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1$

$\Rightarrow 12h-5k-1=\pm 13$

$\Rightarrow 12h-5k=\pm 14$

$\Rightarrow 12h-5k=14,\ 12h-5k=-14$

$\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0$

$\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0$

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0

3 years ago