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kirill115 [55]
3 years ago
6

Brainliest for the correct ANSWER!!! Help ASAP

Mathematics
2 answers:
gladu [14]3 years ago
4 0

Answer:

slope = (1,3)

Step-by-step explanation:


WITCHER [35]3 years ago
3 0

Slope = change in Y over the change in x:

Slope = (3-1) / (3- -3)

Slope = 2 / 6

Slope = 1/3

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Ksivusya [100]

Answer:

three hundred twenty six thousandths

8 0
3 years ago
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Find the slope of the line that passes through (5, 1) and (9, 2).<br> It’s due in 4 minutes
Mekhanik [1.2K]

Answer:

1/4

Step-by-step explanation:

(2-1)/(9-5)

1/4

8 0
3 years ago
13. Solve 5/3x + 1/3x=13 1/3 + 8/3x
N76 [4]

Answer:

The answer: 53x+13x=1313+83x

Step 1: Simplify both sides of the equation.

53x+13x=1313+83x

(53x+13x)=83x+1313(Combine Like Terms)

2x=83x+1313

2x=83x+1313

Step 2: Subtract 8/3x from both sides.

2x−83x=83x+1313−83x

−23x=1313

Step 3: Multiply both sides by 3/(-2).

(3−2)*(−23x)=(3−2)*(1313)

x=−1312

Answer:

x=−131/2

Step-by-step explanation:

7 0
4 years ago
Pls help me ASAP due by 12:00 I’ll brainlest
xxMikexx [17]

Answer:

11x-6

Step-by-step explanation:

not good at explanations ig

6 0
3 years ago
Find the solutions of the following trigonometric equation in the interval [0,2π). tan^2x+secx=1
Citrus2011 [14]
\tan ^2x+\sec x=1

Given the next trigonometric identity:

\begin{gathered} \tan ^2x+1=\sec ^2x \\ \text{ Or} \\ \tan ^2x=\sec ^2x-1 \end{gathered}

Substituting this identity into the equation:

\sec ^2x-1+\sec x=1

Subtracting 1 at both sides of the equation:

\begin{gathered} \sec ^2x-1+\sec x-1=1-1 \\ \sec ^2x+\sec x-2=0 \end{gathered}

Replacing with:

\begin{gathered} y=\sec x \\ y^2=\sec ^2x \end{gathered}

we get:

y^2+y-2=0

Applying the quadratic formula with a = 1, b = 1 and c = -2:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Recalling that y = sec(x), then we have two options:

\begin{gathered} \sec x=1 \\ \text{and} \\ \sec x=-2 \end{gathered}

By definition:

\sec x=\frac{1}{\cos x}

Therefore, the first option is:

\begin{gathered} \frac{1}{\cos x}=1 \\ (\frac{1}{\cos x})^{-1}=1^{-1} \\ \cos x=1 \end{gathered}

In the interval of x [0,2π), the solution to this equation is 0.

Now, considering the second option:

\begin{gathered} \frac{1}{\cos x}=-2 \\ (\frac{1}{\cos x})^{-1}=(-2)^{-1} \\ \cos x=-\frac{1}{2} \end{gathered}

In the interval of x [0,2π), the solutions to this equation are 2π/3 and 4π/3.

In summary, the solutions to tan^2⁡(x) + sec(x) = 1 are:

x=0,\frac{2\pi}{3},\frac{4\pi}{3}

5 0
1 year ago
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