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V125BC [204]
3 years ago
12

Is the bisector of ∠ABC and is the bisector of ∠ACB.

Mathematics
1 answer:
ycow [4]3 years ago
7 0

Answer:

ASA

Step-by-step explanation:

You can show the angles at either end of segment BC in triangles MBC and LCB are congruent, so you have two angles and the segment between. The appropriate theorem in such a case is ASA.

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What is the coefficient of the term 53xy ? A. 5 B.1/3 C.5xy D.5/3
svet-max [94.6K]
It is d i did the test and got an a
7 0
3 years ago
In the expression -3g + 12, how many terms are in the expression? *
IgorC [24]

Answer:

There are 1 terms.

5x+kr has 3 terms

Step-by-step explanation:

-3g has 1 term. 12 has none. so it is 1

5x has 1 term and kr has 2 terms and 2+1 is 3 terms.

7 0
3 years ago
What is the total depreciation (at 10% rate) on a blower purchased 3 years ago for $499?
GrogVix [38]

\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &499\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.10\\ t=\textit{elapsed time}\dotfill &3\\ \end{cases} \\\\\\ A=499(1-0.10)^3\implies A=499(0.9)^3\implies A=363.771 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{total depreciation 499 - 363.771}}{135.229}~\hfill

6 0
3 years ago
A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of
rusak2 [61]

The height of the tunnel 5 feet from the edge is 13.82 feet option second 13.82 feet is correct.

<h3>What is an ellipse?</h3>

An ellipse is a locus of a point that moves in a plane such that the sum of its distances from the two points called focus adds up to a constant. It is taken from the cone by cutting it at an angle.

We have:

A tunnel is constructed with a semi-elliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet.

2a = 60 (width of the tunnel is 60 feet)

a = 30

And the maximum height at the center of the tunnel is 25 feet

b = 25

Let's assume the center of the ellipse is at the origin.

So the equation of the ellipse:

\rm \dfrac{x^2}{30^2}+\dfrac{y^2}{25^2}=1

Now plug x = a - 5 = 30 - 5 = 25

\rm \dfrac{25^2}{30^2}+\dfrac{y^2}{25^2}=1

After solving:

\rm \dfrac{y^2}{25^2}=1-\dfrac{25}{36}

\rm y^2=\dfrac{6875}{36}

y = ±13.819 ≈ ±13.82

Height cannot be negative

y = 13.82 feet

Thus, the height of the tunnel 5 feet from the edge is 13.82 feet option second 13.82 feet is correct.

Learn more about the ellipse here:

brainly.com/question/19507943

#SPJ1

5 0
2 years ago
Solve the following System of Equations<br> 5x+4y-z=1<br> 2x-2y+z=1<br> -x-y+z=2
podryga [215]

Answer:

x = 0 , y = 1 , z = 3

Step-by-step explanation:

Solve the following system:

{5 x + 4 y - z = 1 | (equation 1)

2 x - 2 y + z = 1 | (equation 2)

-x - y + z = 2 | (equation 3)

Subtract 2/5 × (equation 1) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - (18 y)/5 + (7 z)/5 = 3/5 | (equation 2)

-x - y + z = 2 | (equation 3)

Multiply equation 2 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

-x - y + z = 2 | (equation 3)

Add 1/5 × (equation 1) to equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y/5 + (4 z)/5 = 11/5 | (equation 3)

Multiply equation 3 by 5:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x - y + 4 z = 11 | (equation 3)

Subtract 1/18 × (equation 2) from equation 3:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+(65 z)/18 = 65/6 | (equation 3)

Multiply equation 3 by 18/65:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y + 7 z = 3 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 7 × (equation 3) from equation 2:

{5 x + 4 y - z = 1 | (equation 1)

0 x - 18 y+0 z = -18 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 2 by -18:

{5 x + 4 y - z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 4 × (equation 2) from equation 1:

{5 x + 0 y - z = -3 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Add equation 3 to equation 1:

{5 x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 1 by 5:

{x+0 y+0 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Collect results:

Answer: {x = 0 , y = 1 , z = 3

3 0
4 years ago
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