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Arada [10]
2 years ago
8

Alessandro wrote the quadratic equation –6 = x2 + 4x – 1 in standard form. What is the value of c in his new equation?

Mathematics
2 answers:
garri49 [273]2 years ago
8 0
X^2+4x-1+6=0
x^2+4x+5=0
c=5
c is 5
mafiozo [28]2 years ago
8 0

Answer:

The value of c is 5


Step-by-step explanation:

we know that

The standard form of the quadratic equation is equal to

ax^{2} +bx+c=0

In this problem we have

-6=x^{2}+4x-1

Convert to standard form

Equate the quadratic equation to zero

x^{2}+4x-1+6=0

x^{2}+4x+5=0

so


a=1\\b=4\\c=5



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2 years ago
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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
MAVERICK [17]
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
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      = cot2A

This verifies the identity.
4 0
1 year ago
Consider the graph of Bx + Cy = A where A, B, and C are all positive constants. Find the coordinates of the x-intercept. Use upp
timofeeve [1]
That is where the line crosses the x axis or where y=0
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1 year ago
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Given: x ∥ y and w is a transversal Prove: ∠3 ≅ ∠6 Parallel lines x and y are cut by transversal w. On line x where it intersect
SOVA2 [1]

Answer:

The missing reason in the proof is transitive property

Step-by-step explanation:

<u>Statement </u>                                    <u>Reason </u>

1. x ∥ y w is a transversal              1. given

2. ∠2 ≅ ∠3                                    2. def. of vert. ∠s

3. ∠2 ≅ ∠6                                    3. def. of corr. ∠s

4. ∠3 ≅ ∠6                                    4.  ??????????

From the statements 2 and 3

The previous proved statement to make use of the transitive property reason or proof

∴ 4. ∠3 ≅ ∠6                                    4. transitive property

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alukav5142 [94]

Answer:

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* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.

Step-by-step explanation:

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Equation II: 2x + 3y = 2

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Where to Eliminate x :

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* Elimination; a coefficient in Equation I is an integer multiple of a coefficient in Equation II.

* Elimination; a coefficient in Equation II is an integer multiple of a coefficient in Equation I.

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