All categories

3 years ago

What is the area of the square

Mathematics

2 answers:

AysviL [449]3 years ago

6 0

the area of the square would be 49

STALIN [3.7K]3 years ago

3 0

49 inches (7 multiplied by 7)

You might be interested in

Im given a XY line, the midpoint is (3.-5) and the Y end is (0,2) how do I find the X ending of the line

romanna [79]

Using the midpoint formula, the coordinates of the endpoint X, are calculated as: X(6, -12).

<h3>How to Apply the Midpoint Formula?</h3>

The midpoint formula is expressed as: M(x, y) = [(x_1 + x_2)/2, (y_1 + y_2)/2].

We are given the following:

Line XY has a midpoint with the coordinates: (3, -5)

It has one endpoint: Y(0, 2)

To find the coordinates of the other endpoint, X, let:

X(?, ?) = (x_1, y_1)

Y(0, 2) = (x_2, y_2)

M (x, y) = (3, -5)

Plug in the values into the midpoint formula

M(3, -5) = [(x_1 + 0)/2, (y_1 + 2)/2]

Isolate and solve each variable

3 = (x_1 + 0)/2

6 = x_1

x_1 = 6

-5 = (y_1 + 2)/2

-10 = y_1 + 2

-10 - 2 = y_1

y_1 = -12

Therefore, the coordinates of the endpoint X, are: X(6, -12).

Learn more about the midpoint formula on:

brainly.com/question/13115533

#SPJ1

7 0

2 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

3 years ago

In Euclidean geometry, the sum of the lengths of 2 sides of a triangle is greater than the length of the third side. The lengths

bearhunter [10]

Answer: $x \in (\frac{5}{3},\frac{19}{3})$