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3 years ago

Determine the slope from the given graph below:

Mathematics

2 answers:

Yuliya22 [10]3 years ago

5 0

Answer:

y=(10×10)÷(1-2) =N

Step-by-step explanation:

y -is the numerical angle of triangle

10×10because the triangle have a ten angle

1-2 cause the line of triangle

Pie3 years ago

3 0

Answer:

y = -2x - 3

Step-by-step explanation:

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The scatter plot below shows one class of Spanish students' time spent studying for their final versus the grade that they earne

IgorC [24]

<span>If you made a line of best fit, then the score would most likely be 95.</span>

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3 years ago

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The points(1,r) and (5,-4) lie on a line with slope 5. find the missing coordinate r.

Arte-miy333 [17]

The value of r is -24

Step-by-step explanation:

Given points are:

(1,r) and (5,-4)

Let the points be:

(x1,y1) = (1,r)

(x2,y2) = (5,-4)

m = 5

The slope is given by:

$m = \frac{y_2-y_1}{x_2-x_1}\\5 = \frac{-4-r}{5-1}\\5 = \frac{-4-r}{4}\\5 * 4 = -4-r\\20 = -4-r\\20+4 = -4-r+4\\24 = -r\\r = -24$

Hence,

The value of r is -24

Keywords: Slope, coordinates

Learn more about slope at:

#LearnwithBrainly

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3 years ago

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Reil [10]

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3 years ago

What is the amplitude of y = -8sin(3x)?

Sonbull [250]

Answer:

for y=-8sin(3x)

Amplitude = 8

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3 years ago

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Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

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