All categories

2 years ago

Solve for x in the equation x squared + 11 x + StartFraction 121 Over 4 EndFraction = StartFraction 125 Over 4 EndFraction.

Mathematics

2 answers:

Solnce55 [7]2 years ago

8 0

Answer:

the answer is x=-11/2 +or- 5 (rooted)5/2

AKA D

Step-by-step explanation:

Ira Lisetskai [31]2 years ago

6 0

Answer:

Below

Step-by-step explanation:

● x^2 + 11x + 121/4 = 125/4

Substract 125/4 from both sides:

● x^2 + 11x + 121/4-125/4= 125/4 -125/4

● x^2 + 11x - (-4/4) = 0

● x^2 +11x -(-1) = 0

● x^2 + 11 x + 1 = 0

This is a quadratic equation so we will use the determinanant (b^2-4ac)

● a = 1

● b = 11

● c = 1

● b^2-4ac = 11^2-4*1*1 = 117

So this equation has two solutions:

● x = (-b -/+ √(b^2-4ac) ) / 2a

● x = (-11 -/+ √(117) ) / 2

● x = (-11 -/+ 3√(13))/ 2

● x = -0.91 or x = -10.9

Round to the nearest unit

● x = -1 or x = -11

The solutions are { -1,-11}

You might be interested in

Your cousin borrows $1125 to repair her car. The simple annual interest rate is 10%. She makes equal monthly payments of $25. Ho

Elena-2011 [213]

Answer:

neat

Step-by-step explanation:

8 0

2 years ago

Complete the square to rewrite y = x2 - 6x + 15 in vertex form. Then state

love history [14]

Since $x^2$ is the square of x and 6x is twice the product between x and 3, the second square must be 3 squared, i.e. 9.

So, if we think of 15 as 9+6, we have

$x^2-6x+9+6 = (x-3)^2+6$

Which is the required vertex form. This form tells us imediately that the vertex is the point (3,6).

Since the leading coefficient is 1, the parabola is facing upwards (it's U shaped), so the vertex is a minimum.

5 0

2 years ago

Determine the value of $a$. [asy] pair w=(0,4); pair x=(0,0); pair y=(4,0); pair z=y+7/sqrt(2)*(1,1); dot(w); dot(x); dot(y); do

grandymaker [24]

Answer:

a=7

Step-by-step explanation:

The image is rendered and attached below.

Triangle WXY is an Isosceles right triangle, since WX=XY.

First, we determine the length of WY using Pythagoras Theorem.

$WY=\sqrt{4^2+4^2}\\WY=\sqrt{32}$

Since triangle WXY is Isosceles, $\angle XYW=45^\circ$

$\angle XYZ=\angle XYW+\angle WYZ\\135^\circ=45^\circ+\angle WYZ\\\angle WYZ=135^\circ-45^\circ=90^\circ$

Therefore:

Triangle WYZ is a right triangle with WZ as the hypothenuse.

Applying Pythagoras Theorem

$WZ^2=WY^2+YZ^2\\9^2=(\sqrt{32})^2+a^2\\a^2=81-32\\a^2=49\\a^2=7^2\\$Therefore: a=7$

3 0

2 years ago

Solve the following by factoring 4x² + 12 =7x

LiRa [457]

Answer:

4x² + 12 = 7x

4x² + 12 - 7x = 0

use the quadratic formula and get your answer!

Step-by-step explanation:

3 0

2 years ago

Solve the following quadratic equations by extracting square roots.Answer the questions that follow.

Vikentia [17]

Answer:

1. x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

Taking square root on both sides

$\sqrt{x^2}=\sqrt{16}\\\sqrt{x^2}=\sqrt{(4)^2}\\$

x=±4

2. t^2=81

Taking square root on both sides

$\sqrt{t^2}=\sqrt{81}\\\sqrt{t^2}=\sqrt{(9)^2}$

t=±9

3. r^2-100=0

$r^{2}-100=0\\r^2 =100\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{r^2}=\sqrt{100}\\\sqrt{r^2}=\sqrt{(10)^2}$

r=±10

4. x²-144=0

x²=144

Taking square root on both sides

$\sqrt{x^2}=\sqrt{144}\\\sqrt{x^2}=\sqrt{(12)^2}$

x=±12

5. 2s²=50

$\frac{2s^2}{2} =\frac{50}{2}\\s^2=25\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{s^2}=\sqrt{25}\\\sqrt{s^2}=\sqrt{(5)^2}$

s=±5 ..

4 0

1 year ago

Read 2 more answers