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I am Lyosha [343]
1 year ago
10

In the diagram blow, find the measure of angle BCE

Mathematics
1 answer:
Irina-Kira [14]1 year ago
6 0

We have the next

AB is perpendicular with AD

DE is perpendicular with DC

angle B= angle ECD

angle ACD= 56°

We need to remember that the sum of the interior angles of a triangle is 180° and a right triangle is equal to 90°

So we have the next equation for angle B

m\angle B+56+90=180

we isolate the angle B

m\angle B=180-90-56=34m\angle B=34\text{\degree}

then for the angle BCE it forms an angle of 180° with angle BCA and ECD, therefor angle BCE is

m\angle BCE=180-56-34=90\degree

ANSWER

m

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B) Cube

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C) Rectangular pyramid has one rectangle base and 4 congruent triangles

D) Rectangular prism has 3 sets of rectangles; total of 6 rectangles

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3 years ago
Write the equation of a line with slope 3 and y-intercept −3.
natta225 [31]
The equation would be y = 3x-3




4 0
3 years ago
Read 2 more answers
A right-angled triangle has shorter side lengths exactly c^2-b^2 and 2bc units respectively, where b and c are positive real num
yKpoI14uk [10]

Answer: hypotenuse = c^{2} + b^{2}

Step-by-step explanation: Pythagorean theorem states that square of hypotenuse (h) equals the sum of squares of each side (s_{1},s_{2}) of the right triangle, .i.e.:

h^{2} = s_{1}^{2} + s_{2}^{2}

In this question:

s_{1} = c^{2}-b^{2}

s_{2} = 2bc

Substituing and taking square root to find hypotenuse:

h=\sqrt{(c^{2}-b^{2})^{2}+(2bc)^{2}}

Calculating:

h=\sqrt{c^{4}+b^{4}-2b^{2}c^{2}+(4b^{2}c^{2})}

h=\sqrt{c^{4}+b^{4}+2b^{2}c^{2}}

c^{4}+b^{4}+2b^{2}c^{2} = (c^{2}+b^{2})^{2}, then:

h=\sqrt{(c^{2}+b^{2})^{2}}

h=(c^{2}+b^{2})

Hypotenuse for the right-angled triangle is h=(c^{2}+b^{2}) units

3 0
3 years ago
Find an equation of the line parallel to x=5y-9 and passing through (2,-4). Express in standard form.
Anon25 [30]

Answer:

x - 5y = 22

Step-by-step explanation:

Step 1:  rewrite the equation of the given line in to slope-intercept form by solving for y

 x = 5y - 9    

     -5y = -x - 9               (subtract 5y and x from both sides)

        y = x/5 + 9/5           (divide both side by -5)

Step 2:  Our line is parallel to this line, so it has the same slope, but a different  y-intercept, so set up the equation...

  y = x/5 + b      

We are given a  point (x, y) of (2, -4), so plug that in and solve for b.  

 -4 = 2/5 + b

            -4 - 2/5 = b      (subtract 2/5 from both sides to isolate b)  

                   -20/5 - 2/5 = b

 

                         -22/5 = b      (simplify)

So the equation of our line is  y = x/5 - 22/5

Step 3:  Standard form is ax + by = c, where a is a positive integer

subtract x/5 from both sides...

-x/5 + y = - 22/5

multiply by -5 so x becomes a positive integer

x - 5y = 22  

3 0
3 years ago
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