All categories

3 years ago

A student performed the following system to find the solution to the equation x2-2x-8=0

Mathematics

2 answers:

Molodets [167]3 years ago

5 0

The student messed up on step one. The proper factoring is (x-4) (x+2)

Vsevolod [243]3 years ago

3 0

Answer:

B. in step 1

Step-by-step explanation:

Given that a student performed the following steps to find the solution of the quadratic equation $x^2-2x-8=0$ .

Step 1: Factor the polynomial into (x-4) and (x-2).

Step 2: x-4=0 and x-2=0

Step 3: x=4 and x=2.

Now we need to find about which step has the error.

let's multiply (x-4) and (x-2)

we get : $(x-4)(x-2)=x^2-4x-2x+8=x^2-8x+8$

While original problem has -8 as the last term. So that means factors are wrong.

Hence choice B. in step 1. is correct

You might be interested in

T¹/³+t¹/²=12.Pleass solve for me

mart [117]

I think it’s 72/5 there’s the proof I hope im was helpful

6 0

3 years ago

Read 2 more answers

What is the value for x in the diagram? Rounded to the nearest whole degree.

AnnZ [28]

Answer:

Step-by-step explanation:

cos(x)=31/47

x=cos^-1 (31/47)

8 0

3 years ago

Read 2 more answers

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

Find the perimeter of the polygon

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

Brainliest to the correct answer, but answer ASAP

Lyrx [107]

Answer:

B) 25 / 8

Step-by-step explanation:

We know that pi = 3.14*, or "3 and a little bit more". 25/8=3.125 which is approximately pi.

7 0

2 years ago

An Open rectangular tank of length 18 cm and breadth 16 cm contains water to a depth of 13 cm. The water is poured into a cylind

san4es73 [151]

Answer:

Step-by-step explanation

Volume of tank = l* b*h

= 18*16*13

=3744 cubic cm

Volume of cylinder = volume of tank

πr²h = 3744

3.14* (17/2)² * h = 3744

h= 3744 *2 *2/*17*17*3.14

h=16.5 cm

4 0

3 years ago