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STALIN [3.7K]
3 years ago
5

Help me pleaseeeeeeeeeeee

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

similar sss

Step-by-step explanation:

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An item that normally sells for $130 is marked down by 30% what is the sale price
Elis [28]

The answer is 91. Amount saved = Orig. Price x Discount % / 100 , Sale Price = Orig. Price - Amount Saved

6 0
3 years ago
I WILL MARK BRAINLIEST IF YOU ANSWE AND THANK YOU
professor190 [17]

Answer:

1. $106.20

2. $17.73

3. $372.59

4. $57

Step-by-step explanation:

1. Addition 97.95+8.25=106.20

2 multiplication. 1.97x9=17.73

3. Subtraction 447.59-75.00=372.59

4. Division. 285/5=57

6 0
3 years ago
Read 2 more answers
Ms. Zerngast’s favorite candy is a snickers bar. It costs 7.80$ for 6 large snickers candy bars. She likes to buy them at our lo
Leto [7]
14

If 7.80 buys 6 chocolates then 15.60 should buy 12 with a remainder of 2.60
7.80 divided by 6 is 1.30 and 1.30 +1.30 = 2.60 so 12 +12 = 14
3 0
3 years ago
Los lados de un triángulo rectángulo miden 6m, 8m y 10m. respectivamente. ¿Cuánto medirán los catetos de un triángulo semejante
OlgaM077 [116]

Answer: Los catetos medirán 12m y 9m.

Step-by-step explanation:

Bueno, sabemos que los lados de nuestro triangulo son 6m, 8m y 10m.

Podemos usar relaciones trigonométricas para encontrar los ángulos de este triangulo.

Cos(X) = cateto adyacente/hipotenusa

Si tomamos al cateto adyacente como 8m, tenemos:

Cos(X) = 8m/10m

X = Acos(8/10)  

Esto nos dice que para que nuestro otro triangulo rectángulo sea equivalente al primero, entonces los nuevos catetos C1  y C2 tienen que ser tal que los cocientes entre los catetos y la hipotenusa se mantengan constantes:

C1/15m = 8m/10m

C1 = (8/10)*15m = 12m

C2/15m = 6m/10m  

C2 = (6/10)*15m = 9m.

5 0
4 years ago
Please help me prove that ED is congruent to BA! I thought I had it right by proving right and vertical angles, but I’m missing
Vikki [24]

Step-by-step explanation:

\overline{DA} \text{ bisects } \overline{EB}     is given

\overline{BC} \cong \overline{EC}  definition of <u>bisect</u>

<u />\overline{EB} \perp \overline{ED}, \, \overline{EB} \perp \overline{BA}  are both given

<u />\angle B, \, \angle E \text{ are right angles}  definition of perpendicular

\angle B \cong \angle E   because all right angles are congruent

\angle{ACB} \cong \angle{DCE}  vertical angles are congruent

\triangle{ACB} \cong \triangle{DCE}   ASA (angle-side-angle)

\overline{ED} \cong \overline{BA}  CPCTC (corresponding parts of congruent triangles are congruent)

5 0
3 years ago
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