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melomori [17]
3 years ago
5

Solve each problem and show the work for each one.

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
8 0

Answer:

Solve by factoring.

1. 2x^2+9x+26=-4x+5

2x²+9x+4x+26-5=0

2x²+13x+21=0

doing middle term factorization

2x²+7x+6x+21=0

x(2x+7)+3(2x+7)=0

<u>(</u><u>2</u><u>x</u><u>+</u><u>7</u><u>)</u><u>(</u><u>x</u><u>+</u><u>3</u><u>)</u><u>=</u><u>0</u>

<u>either</u>

<u>x</u><u>=</u><u>-</u><u>7</u><u>/</u><u>2</u>

<u>or</u>

<u>x</u><u>=</u><u>-</u><u>3</u><u>.</u>

2.

3x^2-6x-4=0

comparing above equation with ax²+bx+c=0

we get

a=3

b=-6

c=-4

By using quadratic equation

x=\frac{ - b±\sqrt{ {b}^{2} - 4ac } }{2a}

Substituting value

x=\frac{ 6±\sqrt{ {-6}^{2} - 4*3*-4 }}{2*-4}

x=\frac{ 6±\sqrt{ {84}}}{-8}

x=\frac{ 6±2\sqrt{ {21}}}{-8}

-8x=6±2√21

taking positive

-8x=6+2√21

x=\frac{6+2√21}{-8}

x=-\frac{3+√21}{4}

taking negative

-8x=6-2√21

x=\frac{6-2√21}{-8}

x=-\frac{3-√21}{4}

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FrozenT [24]

Answer:

I'm sorry I don't have answer but.

it's not C!

8 0
2 years ago
Prove 2/sqrt3cosx+sinx=sec(pi/6-x)
dlinn [17]

Thus L.H.S = R.H.S that is 2/√3cosx + sinx  = sec(Π/6-x) is proved

We have to prove that

2/√3cosx + sinx  = sec(Π/6-x)

To prove this we will solve the right-hand side of the equation which is

R.H.S = sec(Π/6-x)

          = 1/cos(Π/6-x)

[As secƟ = 1/cosƟ)

           = 1/[cos Π/6cosx + sin Π/6sinx]

[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]

           = 1/[√3/2cosx + 1/2sinx]

            = 1/(√3cosx + sinx]/2

            = 2/√3cosx + sinx

    R.H.S = L.H.S

Hence 2/√3cosx + sinx  = sec(Π/6-x) is proved

Learn more about trigonometry here : brainly.com/question/7331447

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5 0
1 year ago
Alon was curious if segments \overline{JK} JK start overline, J, K, end overline and \overline{NM} NM start overline, N, M, end
BartSMP [9]

The answer is A because Alon could have rotated the figure and discovered that they are congruent

6 0
3 years ago
Please help me with this algebra PLEEASSEee!!!
amid [387]
11. 5x^2-30=0
+30 to both sides
5x^2=30
÷5 both sides
x^2=6
square root both sides
x= +square root 6 ,- square root 6

12. 4x^2+10=26
-10 both sides
4x^2=16
÷4 both sides
x^2=4
square root both sides
x=+2,-2

13. a=72yrd^2
l=2w
w=?
a=l×w
72=2w×w
72=2w^2
÷2 both sides
36=w^2
square root both sides
w=+6,-6
a measurement can't be - so w=6
Plug into l=2w
l=2×6

l=12
w=6
5 0
3 years ago
I need help please, I dont know how to do this
Oliga [24]

Answer:

Step-by-step explanation:

Odd numbers backwards - 9, 7, 5, 3

1 = -2+9=7

2 = -1+7=6

3 = 0+5=5

4 = 1 +3=4

3 0
3 years ago
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