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3 years ago

Helppp ME PLEASEE (10 points )

Mathematics

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer:

obtuse 9 acute 10

Step-by-step explanation:

obtuse is under 30 degrese and acute is above

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4 years ago

Find the distance between the two points. A(0,1) and B( 6, 3.5)

Karo-lina-s [1.5K]

Answer:

AB = 6.5

Step-by-step explanation:

If you use the distance formula, you can find the answer.

$\begin{array}{l}AB=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\=\sqrt{\left(6-0\right)^2+\left(3.5-1\right)^2}\\=\sqrt{6^2+2.5^2}\\=\sqrt{36+6.25}\\=\sqrt{42.25}\\AB=6.5\end{array}$

Therefore, the distance between the points A(0, 1) and B(6, 3.5) is 6.5. Hope this helps!

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3 years ago

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Give the equation of a line that goes through the point ( − 21 , 2 ) and is perpendicular to the line 7 x − 4 y = − 12 . Give yo

nlexa [21]

Given:

Equation of line $7x-4y=-12$ .

To find:

The equation of line that goes through the point ( − 21 , 2 ) and is perpendicular to the given line.

Solution:

The given equation of line can be written as

$7x-4y+12=0$

Slope of line is

$\text{Slope}=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}$

$m_1=-\dfrac{7}{(-4)}$

$m_1=\dfrac{7}{4}$

Product of slopes of two perpendicular lines is -1. So, slope of perpendicular line is

$m_1m_2=-1$

$m_2=-\dfrac{1}{m_1}$

$m_2=-\dfrac{4}{7}$ $[\because m_1=\dfrac{7}{4}]$

Now, the slope of perpendicular line is $m_2=\dfrac{4}{7}$ and it goes through (-21,2). So, the equation of line is

$y-y_1=m_2(x-x_1)$

$y-2=-\dfrac{4}{7}(x-(-21))$

$y-2=-\dfrac{4}{7}x-\dfrac{4}{7}(21)$

$y-2=-\dfrac{4}{7}x-12$

$y=-\dfrac{4}{7}x-12+2$

$y=-\dfrac{4}{7}x-10$

Therefore, the required equation in slope intercept form is $y=-\dfrac{4}{7}x-10$ .

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3 years ago

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Yakvenalex [24]

Answer:

192

Step-by-step explanation:

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2 years ago

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