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3 years ago

What is the measure of the missing angle?

Mathematics

2 answers:

Hoochie [10]3 years ago

8 0

The answer would be D because triangles add up to 180 degrees and the two angles together only add up to 140, so the missing angles would be 40 degrees

sergiy2304 [10]3 years ago

4 0

D. 40 degrees

This is the answer because 70+70=140
There is 180 degrees in a Triangle. So you must subtract 140 from 180.
180-140=40.

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A quadrilateral is graphed in the coordinate plane below. Which classification best describes the quadrilateral (parallelogram,

Ahat [919]

Answer:

Trapezoid

Step-by-step explanation:

Given quadrilateral has vertices at points A(-2,-1), B(3,13), C(15,5) and D(13,-11).

Find slopes of lines AD and BC:

$\text{Slope}_{AD}=\dfrac{y_D-y_A}{x_D-x_A}=\dfrac{-11-(-1)}{13-(-2)}=\dfrac{-11+1}{13+2}=\dfrac{-10}{15}=-\dfrac{2}{3}\\ \\\text{Slope}_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{5-13}{15-3}=\dfrac{-8}{12}=-\dfrac{2}{3}$

Since the slopes are the same, lines AD and BC are parallel.

Find slopes of lines ABD and CD:

$\text{Slope}_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{13-(-1)}{3-(-2)}=\dfrac{14}{5}\\ \\\text{Slope}_{CD}=\dfrac{y_D-y_C}{x_D-x_C}=\dfrac{-11-5}{13-15}=\dfrac{-16}{-2}=8$

Since the slopes are different, lines AB and CD are not parallel.

This means quadrilateral ABCD is trapezoid (two opposite sides - parallel and two another opposite sides - not parallel)

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3 years ago

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Step-by-step explanation:

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Step-by-step explanation:

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Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the give

frutty [35]

Answer:

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

Step-by-step explanation:

to find the tangent line we need to find the derivative of the function g(x).

$g(x) =e^{x^3}$

we know that $\frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)$

$g'(x) =e^{x^{3}}(3 x^{2})$

$g'(x) =3 x^{2} e^{x^{3}}$

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

$m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}$

using the equation of line:

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

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

$(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\$

$y=\dfrac{3x}{e}+\dfrac{4}{e}$

this is the equation of the tangent at point (-1,1/e)

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3 years ago