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4 years ago

Plz explain and prove the triangles congruence.

Mathematics

1 answer:

Ad libitum [116K]4 years ago

5 0

Answer:

(3) ∠BCA ≅ ∠DAC

Step-by-step explanation:

BC and AD are parallel. AC is a transversal line passing through both lines. That means ∠BCA and ∠DAC are alternate interior angles. Therefore, they are congruent.

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Joe wants to rent an apartment with an initial monthly rent of $1,400. He has been told that the landlord raises the rent 1.25%

nadya68 [22]

Answer: his pay for the 4th year is $1453.16

Step-by-step explanation:

The landlord raises the rent 1.25% each year. It means that the rent is increasing in geometric progression.

The formula for determining the nth term of a geometric progression is expressed as

Tn = ar^(n - 1)

Where

a represents the first term of the sequence.

r represents the common ratio.

n represents the number of terms.

From the information given,

a = $1,400

r = 1 + 1.25/100 = 1.0125

n = 4 years

The 4th term(year), T4 is

T4 = 1400 × 1.0125^(4 - 1)

T4 = 1400 × 1.0125^3

T4 = $1453.16

8 0

3 years ago

What is -87 = 1 + 8x?

Leno4ka [110]

Answer:

-87= 1 + 8x

-88= 8x

x= -11

Basically bring over the one to the other side of the equation and solve for x by dividing.

6 0

3 years ago

PLEASE HELP!!

Bezzdna [24]

Answer:

$y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

$y=ax^2+b$

This parabola passes through the point R(3,9), then

$9=a\cdot 3^2+b\\ \\9=9a+b$

The area of the right triangle PQR is

$A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR$

Find PQ and QR, if $P(x_1,0),\ Q(3,0),\ R(3,9):$

$PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9$

Now,

$40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7$

We get two possible points $P_1(-7,0)$ and $P_2(13,0).$

For point $P_1:\\$

$0=a\cdot (-7)^2+b\\ \\49a+b=0$

So,

$b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}$

For point $P_2:\\$

$0=a\cdot (13)^2+b\\ \\169a+b=0$

So,

$b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

5 0

3 years ago

Please answer and i have to show work

stealth61 [152]

Answer:

Step-by-step explanation:

Calculate the slope m using the slope formula and equate to $\frac{3}{2}$

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (3, 2) and (x₂, y₂ ) = (r, - 4)

m = $\frac{-4-2}{r-3}$ = $\frac{-6}{r-3}$ , then equating gives

$\frac{-6}{r-3}$ = $\frac{3}{2}$ ( cross- multiply )

3(r - 3) = - 12 ( divide both sides by 3 )

r - 3 = - 4 ( add 3 to both sides )

r = - 1 → B

4 0

3 years ago

Write an equivalent expression to the following 6(b-2)

NeX [460]

The answer is D............

6 0

3 years ago