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2 years ago

How do you do pemdas? Please help, I'm trying to teach my little brother and I forget

Mathematics

2 answers:

Fiesta28 [93]2 years ago

8 0

Answer:

Step-by-step explanation:

PEMDAS is a set of rules which are followed while solving mathematical expressions. These rules start with Parentheses, and then operations are performed on the exponents or powers. Next, we perform operations on multiplication or division from left to right. Finally, operations on addition or subtraction are performed from left to right.

dimaraw [331]2 years ago

8 0

Answer:

PEMDAS is parenthesis, exponent, multiplication, division, addition, and subtraction

Step-by-step explanation:

before anything, you have to check for anything in a parenthesis. If nothing check for any exponents. If nothing do multiplication, and then do all division (of course wherever necessary). After that, you do all of the addition, then subtraction. everything in this order

You might be interested in

Calculate the areas of these rectangles

kykrilka [37]

Answer:

A(a) = 24 cm²

A(b) = 40.04 cm² ≈ 40 cm²

Step-by-step explanation:

Hi there !

area formula (rectangle)

A = l × w

A(a) = 8cm × 3cm = 24 cm

A(b) = 5.2cm × 7.7cm = 40.04 cm² ≈ 40 cm²

Good luck !

8 0

3 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−4/7x, and the line perpendicular to f(x) that passes through

Setler79 [48]

ANSWER:

The area of the triangle bounded by the y-axis is $\frac{7938}{4225} \sqrt{65} \text { unit }^{2}$

SOLUTION:

Given, $f(x)=9-\frac{-4}{7} x$

Consider f(x) = y. Hence we get

$f(x)=9-\frac{-4}{7} x$ --- eqn 1

$y=9-\frac{4}{7} x$

On rewriting the terms we get

4x + 7y – 63 = 0

As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.

Area of right angle triangle = $\frac{1}{ab}$ where a, b are lengths of sides other than hypotenuse.

So, we need find length of f(x) and its perpendicular line.

First let us find perpendicular line equation.

Slope of f(x) = $\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}$

So, slope of perpendicular line = $\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}$

Perpendicular line is passing through origin(0,0).So by using point slope formula,

$y-y_{1}=m\left(x-x_{1}\right)$

Where m is the slope and $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$

$y-0=\frac{7}{4}(x-0)$

$y=\frac{7}{4} x$ --- eqn 2

4y = 7x

7x – 4y = 0

now, let us find the vertices of triangle, one of them is origin, second one is point of intersection of y-axis and f(x)

for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

7y = 63

y = 9

Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

$\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}$

Put x value in (2)

$\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}$

So, the point of intersection is $\left(\frac{252}{65}, \frac{441}{65}\right)$

Length of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right)$ and (0,9)

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}$

Now, length of perpendicular of f(x) is distance between $\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)$

$\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}$