What's the area of a rectangle measuring 13 in. × 12 in

Answers

Answer 1

Answer: it is 156 in.
hope it helped you :)

Related Questions

Me pueden ayudar porfa
Round each decimal number to the nearest integer 11.90
Question 1(04.03 LC)Which of the following options is an equivalent function to f(x) = 2(5)2X? (2 points)a. f(x) = 50xb. f(x) = 100xc. f(x) = 2(25)d. f(x) = 4(25)
HELP ASP!!!!!! The explicit rule for a sequence and one of the specific terms is given. Find the position of the giventerm.f(n) = 1.25n +3.75; 2020 is theth term.
a container was 1/2 filled with water. When 200 ml of water was poured out, it became 1/3 full. Find the capacity of the container in milliliters.

Leo is running in a 5-kilometer race along a straight path. if he is at the midpoint of the path, how many kilometers does he have left to run?

Answers

Let us assume Leo starts from a point (0,0)

Since the total length of the race is 5 km, he will be at (5,0) at the end of the race.

It is given that, he is at the mid point of the race.

Mid point formula is : $((x_(1) +x_(2) ))/(2) , ((y_(1) +y_(2) ))/(2)$

Present coordinate : $(5+0)/(2) , (0+0)/(2) = (2.5,0)$

He still needs to run 5-2.5 = 2.5 km

Suppose that you are a manager of a clothing store and have just purchased 100 shirts for $15 each. After a month of selling the shirts at the regular price, you plan to have a sale giving 30% of the original selling price. However ,you want to make a profit of $6 on each shirt at the sale price. What should you price the shirts at initially to ensure this?

Answers

He should set the original price at $27.30

When you say "Plan to have a sale giving 30% of the original selling price", I suppose it's taking away 30%? So taking away 30% is like $15 - $4.5 = $10.50
Adding $6 more is $16.50.
If you're saying the manager is selling the price 30% of $15 then the price is $4.50 Plus the $6, it should be $10.50

So the answer should either be $16.50 or $10.50
Hope this helps! :)

4. Find the range of f(x) = -x + 4 for the domain {-3, -2, -1, 1}.O {7, 6, 5, 3}
O {-7, -6, -5, -4}
O {7, 6, 5,4}
O {-7, -6, -5, 3}

Answers

It’s this first one- 7,6,5,3
You can get this by plugging each of the x values into the expression -x+4
Hope that helps

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?f(x) = 4x5 – 3x
1 root
2 roots
4 roots
5 roots

Answers

Answer:

Option 4 is correct that is according to fundamental theroem of algebra the given polynomial function will have 5 roots

Step-by-step explanation:

The fundamental theorem of algebra says that the polynomial will have the number of roots equal to the degree of the polynomial.

Degree is the highest power of polynomial.

Here we have given the polynomial $4x^5-3x$ where degree is 5.

Hence, the given polynomial will have 5 roots.

Therefore, option 4 is correct that is 5 roots

Hello,

I say one more time in C there are 5 roots.
$4x^5-3x=x(4x^4-3)\n =x(2x^2+ √(3) )(2x^2- √(3) )\n =x( √(2)x- \sqrt[4]{3} ) ( √(2)x+ \sqrt[4]{3} ) ( √(2)x- i\sqrt[4]{3} ) ( √(2)x+i \sqrt[4]{3} )$

Pentagon PQRST and its reflection, pentagon P'Q'R'S'T', are shown in the coordinate plane below:Pentagon PQRST and pentagon P prime Q prime R prime S prime T prime on the coordinate plane with ordered pairs at P negative 4, 6, at Q negative 7, 4, at R negative 6, 1, at S negative 2, 1, at T negative 1, 4, at P prime 4, 6, at Q prime 7, 4, at R prime 6, 1, at S prime 2, 1, at T prime 1, 4.

What is the line of reflection between pentagons PQRST and P'Q'R'S'T'?

x = 0
y = x
y = 0
x = 1

Answers

I have graphed the given coordinates of both pentagons. The reflection was across the y-axis where coordinates of Pentagon PQRST (-x,y) resulted to Pentagon P'Q'R'S'T' (x,y). The line of reflection between the pentagons was x = 0. Line of reflection is the midway between the pre-image and its reflection.

Yes, the answer is A. x=0. Another way to say "reflect across the y-axis" is to say "reflect across the line x=0" since the line created by graphing x=0 is the same as the y-axis. An image that is a reflection across the y-axis, or across the line x=0, will have opposite x-coordinates from the pre-image but identical y-coordinates. In other words, the x value points just changed signs from negative to positive values since it went through the rule (x, y) → (−x, y).

How would I solve 16x^4-41x^2+25=0 ???

Answers

$16{ x }^( 4 )-41{ x }^( 2 )+25=0$

${ x }^( 4 )={ ({ x }^( 2 )) }^( 2 )\n \n 16{ ({ x }^( 2 )) }^( 2 )-41{ x }^( 2 )+25=0$

First of all to make our equation simpler, we'll equal $x^(2)$ to a variable like 'a'.

So,

${ x }^( 2 )=a$

Now let's plug $x^(2)$ 's value (a) into the equation.

$16{ ({ x }^( 2 )) }^( 2 )-41{ x }^( 2 )+25=0\n \n { x }^( 2 )=a\n \n 16{ (a) }^( 2 )-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (fourth-degree equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^( 2 )-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :

$t{ x }^( 2 )+bx+c=0$

In our equation,

t=16 , b=-41 and c=25

Let's plug the values in the formula to solve.

$t=16\quad b=-41\quad c=25\n \n \frac { -(-41)\pm \sqrt { -(41)^( 2 )-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \n \n \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \n \n \frac { 41\pm \sqrt { 81 } }{ 32 } \n \n \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \n \n \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\n \n a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^( 2 )=a$

So after we found a's solution set, that means.

${ x }^( 2 )=\frac { 50 }{ 32 }$

and

${ x }^( 2 )=1$

We'll also solve this equations to find x's solution set :)

${ x }^( 2 )=\frac { 50 }{ 32 } \n \n \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \n \n { x }^( 2 )=\frac { 25 }{ 16 } \n \n \sqrt { { x }^( 2 ) } =\sqrt { \frac { 25 }{ 16 } } \n \n x=\quad \pm \frac { 5 }{ 4 }$

${ x }^( 2 )=1\n \n \sqrt { { x }^( 2 ) } =\sqrt { 1 } \n \n x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

16x⁴ - 41x² + 25 = 0
16x⁴ - 16x² - 25x² + 25 = 0
16x²(x²) - 16x²(1) - 25(x²) + 25(1) = 0
16(x² - 1) - 25(x² - 1) = 0
(16x² - 25)(x² - 1) = 0
(16x² + 20x - 20x - 25)(x² - x + x - 1) = 0
(4x(4x) + 4x(5) - 5(4x) - 5(5))(x(x) - x(1) + 1(x) - 1(1)) = 0
(4x(4x + 5) - 5(4x + 5))(x(x - 1) + 1(x - 1)) = 0
(4x - 5)(4x + 5)(x + 1)(x - 1) = 0
4x - 5 = 0 or 4x + 5 = 0 or x + 1 = 0 or x - 1 = 0
    + 5 + 5    - 5 - 5 - 1 - 1 + 1 + 1
   4x = 5 4x = -5 x = -1    x = 1
      4    4    4 4
   x = 1¹/₄ x = -1¹/₄

The solution set is equal to {(1¹/₄, -1¹/₄, -1, 1)}.

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