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3 years ago

8

If ∆JKL ≅ ∆PQR, which one of these is not a pair of corresponding parts? ∠J and ∠P and ∠L and ∠R and

1 answer:

Mandarinka [93]3 years ago

8 0

When trying to prove that two triangles are congruent there are three congruency rules SAS, ASA and SSS. From the looks of things though I would say it would be the third one because just from the way you named the triangles I could tell that the two angles could correspond.

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1)Circle 1 is centered at (−4, 5)(−4, 5) and has a radius of 2 centimeters. Circle 2 is centered at (2, 1)(2, 1) and has a radiu

galina1969 [7]

Figures of same shape and size are similar .Two circles C1&C2 will be similar.

Circle 1 has a center of (-4,5) and circle 2 has a center of (2,1) .The x of the center is having the translation x+6 and the y is having a translation of y-4.The center of the circle is dilated by 3 units.

The circles are similar because you can translate Circle 1 using the transformation rule (x+6,y-4 ) and then dilate it using a scale factor of 3.

2) Area of sector = $\pi.r^{2}\alpha$ ÷360.

Where α is the angle made at center.

Area of given sector= π(12)(12)(60)÷360 =24π.

8 0

3 years ago

The options are<br> 40<br> 80<br> 120<br> 140

Nataly [62]

<h2>Here we go ~ </h2>

According to given figure,

$\qquad \sf \dashrightarrow \: \angle ABC + 60° = 180°$

$\qquad \sf \dashrightarrow \: \angle ABC = 180 \degree - 60 \degree$

[ By linear pair ]

$\qquad \sf \dashrightarrow \: \angle ABC = 120 \degree$

now, we can see that :

$\qquad \sf \dashrightarrow \: \angle ABC + \angle BAC = \angle 1$

[ By Exterior angle property of Triangle ]

$\qquad \sf \dashrightarrow \: \angle 1 = 20 \degree + 120 \degree$

$\qquad \sf \dashrightarrow \: \angle 1 = 140 \degree$

4 0

2 years ago

Which shows another way to write 9.54?

Lelu [443]

Answer:

It would be C

Step-by-step explanation:

3 0

3 years ago

Jon has $45.00. He plans to spend 4/5 of his money on sports equipment. How much will he spend

stich3 [128]

He will spend 80% of his money

5 0

3 years ago

I am trying to figure this question: <br><br> 1 + tan^2A = Sec^2A

chubhunter [2.5K]

Answer:

see explanation

Step-by-step explanation:

Using the Pythagorean identity

cos²A + sin²A = 1 ( divide terms by cos²A )

$\frac{cos^2A}{cos^2A}$ + $\frac{sin^2A}{cos^2A}$ = $\frac{1}{cos^2A}$ , that is

1 + tan²A = sec²A ← as required

3 0

3 years ago