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strojnjashka [21]
2 years ago
13

The first step in determining the solution to the system of equations, y = -x2 - 4x - 3 and y = 2x + 5, algebraically is to set

the
two equations equal as X? - 4x - 3 = 2x + 5. What is the next step?
Set y = 0in y = -x? - 4x - 3.
Factor each side of the equation.
Use substitution to create a one-variable equation.
Combine like terms onto one side of the equation
Mathematics
1 answer:
sweet-ann [11.9K]2 years ago
6 0

Answer:

Combine like terms onto one side of the equation

Step-by-step explanation:

You have

-x^2  -4x  - 3  = 2x + 5

It seems that we should combine like terms to make this equations smaller.

-x^2 - 4x - 3  - 2x  - 5 = 0

-x^2  -6x  - 8 = 0

- (x + 2)(x + 4) = 0

x = -2  or  x  = -4

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2 years ago
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Zelda made 5 gallons of lemonade and gave each of her friends an equal amount. Write an expression that represents how much lemo
enot [183]

Answer:

2.5

Step-by-step explanation:

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8 0
3 years ago
First question, thanks. I believe there should be 3 answers
zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

3 0
1 year ago
graham's monthly bank statement showed the following deposists and withdrawals: -$25.20,$52.75,-$22.04,-$8.50,$94.11 if graham's
dezoksy [38]
Answer= $138.98

47.86-25.20+52.75-22.04-8.50+94.11=$138.98
3 0
3 years ago
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