All categories

3 years ago

Solve the expression one half x (48 ÷ 6) − 2 + 5 using PEMDAS. 7 8 9 10

Mathematics

1 answer:

stellarik [79]3 years ago

4 0

7 is the answer

………..

You might be interested in

Perform the following

antoniya [11.8K]

The appropriate number of significant figure of the mathematical operation 16.2156 +0.014 is 16.230.

<h3 /><h3>Significant figures:</h3>

16.2156 is a 6 significant figures number
0.014 is a 2 significant figures number

Therefore,

16.2156 + 0.014 = 16.2296

Therefore, let's round it up to 5 significant figures as follows;

16.2296 = 16.230.

learn more on mathematical operation here: brainly.com/question/13055274?referrer=searchResults

6 0

3 years ago

Find the distance between the lines with the given equations.<br> 3x+y = 1<br> y+6=-3x

den301095 [7]

Answer:

2.21 units

--------------------------------

Given lines:

3x + y = 1 and y + 6 = - 3x

Convert both equations to slope-intercept form of y = mx + b:

3x + y = 1 ⇒ y = - 3x + 1
y + 6 = - 3x ⇒ y = - 3x - 6

Slopes are same, so these are parallel lines, and the distance between parallel lines is:

$d=\cfrac{|b_1-b_2|}{\sqrt{1+m^2} }$ , where b₁, b₂ - y-intercepts, m- slope

Substitute and calculate:

$d=\cfrac{|1+6|}{\sqrt{1+(-3)^2} }=\cfrac{7}{\sqrt{10} } =2.21$

6 0

1 year ago

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.

xz_007 [3.2K]

Answer:

Part 1) $f(x)=\frac{2x-1}{x+2}$ -------> $f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) $f(x)=\frac{x-1}{2x+1}$ -------> $f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) $f(x)=\frac{2x+1}{2x-1}$ -----> $f^{-1}(x)=\frac{x+1}{2(x-1)}$

Part 4) $f(x)=\frac{x+2}{-2x+1}$ ----> $f^{-1}(x)=\frac{x-2}{2x+1}$

Part 5) $f(x)=\frac{x+2}{x-1}$ -------> $f^{-1}(x)=\frac{x+2}{x-1}$

Step-by-step explanation:

Part 1) we have

$f(x)=\frac{2x-1}{x+2}$

Find the inverse

Let

y=f(x)

$y=\frac{2x-1}{x+2}$

Exchange the variables x for y and t for x

$x=\frac{2y-1}{y+2}$

Isolate the variable y

$x=\frac{2y-1}{y+2}\\ \\ xy+2x=2y-1\\ \\xy-2y=-2x-1\\ \\y[x-2]=-2x-1\\ \\y=\frac{-2x-1}{x-2}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) we have

$f(x)=\frac{x-1}{2x+1}$

Find the inverse

Let

y=f(x)

$y=\frac{x-1}{2x+1}$

Exchange the variables x for y and t for x

$x=\frac{y-1}{2y+1}$

Isolate the variable y

$x=\frac{y-1}{2y+1}\\ \\2xy+x=y-1\\ \\2xy-y=-x-1\\ \\y[2x-1]=-x-1\\ \\y=\frac{-x-1}{2x-1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) we have

$f(x)=\frac{2x+1}{2x-1}$

Find the inverse

Let

y=f(x)

$y=\frac{2x+1}{2x-1}$

Exchange the variables x for y and t for x

$x=\frac{2y+1}{2y-1}$

Isolate the variable y

$x=\frac{2y+1}{2y-1}\\ \\2xy-x=2y+1\\ \\2xy-2y=x+1\\ \\y[2x-2]=x+1\\ \\y=\frac{x+1}{2(x-1)}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x+1}{2(x-1)}$

Part 4) we have

$f(x)=\frac{x+2}{-2x+1}$

Find the inverse

Let

y=f(x)

$y=\frac{x+2}{-2x+1}$

Exchange the variables x for y and t for x

$x=\frac{y+2}{-2y+1}$

Isolate the variable y

$x=\frac{y+2}{-2y+1}\\ \\-2xy+x=y+2\\ \\-2xy-y=-x+2\\ \\y[-2x-1]=-x+2\\ \\y=\frac{-x+2}{-2x-1} \\ \\y=\frac{x-2}{2x+1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x-2}{2x+1}$

Part 5) we have

$f(x)=\frac{x+2}{x-1}$

Find the inverse

Let

y=f(x)

$y=\frac{x+2}{x-1}$

Exchange the variables x for y and t for x

$x=\frac{y+2}{y-1}$

Isolate the variable y

$x=\frac{y+2}{y-1}\\ \\xy-x=y+2\\ \\xy-y=x+2\\ \\y[x-1]=x+2\\ \\y=\frac{x+2}{x-1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x+2}{x-1}$

7 0

3 years ago

Find the quadratic equation for y=x^2+6x-7

LenKa [72]

Answer:

x = -1 or x = 7

Step-by-step explanation:

From my understanding, y will be zero and you have to find the value of x.

Step 1: Middle term break

y = x² + 6x - 7

x² + 6x - 7 = 0

x² - 7x + x - 7 = 0

Step 2: Solve

x(x-7) + 1(x-7) = 0

(x+1) (x-7) = 0

x = -1 or x = 7

Therefore, the value of x is either -1 or 7.

5 0

3 years ago

A counterexample for the expression sin y tan y= cos y is 0.

umka21 [38]

Answer:

True

Step-by-step explanation:

To answer this question we must evaluate

y = 0° on both sides of the equation.

For the left side we have:

$sin(0\°) tan(0\°)$

We know that $tan(0\°) =\frac{sin(0\°)}{cos(0\°)}$

We know that $cos(0\°) = 1$ and $sin(0\°) = 0$ .

Therefore $tan(0\°) = 0$ .

Then the left-hand side of the equals is equal to zero.

On the right side we have:

$cos(y)$

When evaluating $cos(y)$ at $y = 0$

We have to $cos(0\°) = 1$ .

0 ≠ 1

The equation is not satisfied. Therefore y = 0 ° is a counterexample to the equation

3 0

3 years ago

Read 2 more answers