PLSSS HELP ME WITH THIS QUESTION!!!!!

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

4 years ago

The tickets for a dance recital cost $5.00 for adults and $2.00 for children. If the total number of tickets sold was 295 and th

dem82 [27]

To solve, you would need to make two equations and substitute one in for the other. Let a = # of adults, and c = # of children

5a + 2c = 1,220
a + c = 295

a + c = 295
-c -c
a = 295 - c

5(295 - c) + 2c = 1,220
1,475 - 5c + 2c = 1,220
1,475 - 3c = 1,220

1,475 - 3c = 1,220
-1,475 -1,475
-3c = -255

-3c/-3 = -255/-3
c = 85

a + c = 295
a + 85 = 295
-85 -85
a = 210

210 adults attended the dance recital

Hope this helps!

6 0

3 years ago

If sin theta = -0.7660, which of the following represents an approximate value of tan theta, for 180 degrees < theta < 270

Elis [28]

We have that
sin ∅=-0.7660

the sin ∅ is negative
so
∅ belong to the III or IV quadrant
but
180°< ∅ < 270°
hence
∅ belong to the III quadrant
sin ∅=-0.7660
sin² ∅+cos² ∅=1
cos² ∅=1-sin² ∅------> cos² ∅=1-(0.7660)²-----> 0.4132
cos ∅=√0.4132-----> cos ∅=0.6428
the value of cos ∅ is negative-------> III quadrant
cos ∅=-0.6428

tan ∅=sin ∅/ cos ∅----> tan ∅=-0.7660/-0.6428----> tan ∅=1.1917

the answer is
tan ∅ is 1.1917

4 0

3 years ago

Q.23 Prove that: (sin4A - cos4A + 1) = 2sin2A

FrozenT [24]

Answer:

Step-by-step explanation:

I think you have the question incomplete, and that this is the complete question

sin^4a + cos^4a = 1 - 2sin^2a.cos^2a

To do this, we can start my mirroring the equation.

x² + y² = (x + y)² - 2xy,

This helps us break down the power from 4 to 2, so that we have

(sin²a)² + (cos²a)² = (sin²a + cos²a) ² - 2(sin²a) (cos²a)

Recall from identity that

Sin²Φ + cos²Φ = 1, so therefore

(sin²a)² + (cos²a)² = 1² - 2(sin²a) (cos²a)

On expanding the power and the brackets, we find that we have the equation proved.

sin^4a + cos^4a = 1 - 2sin^2a.cos^2a

4 0

3 years ago

Write the equation of the line in slope-intercept form when given m=-6 and (-1,3)

Dmitry_Shevchenko [17]

Step-by-step explanation:

y=mx+b

y=-6x+b

3=-6(-1)+b

b=-3

y=-6x-3

3 0

3 years ago