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sergij07 [2.7K]
3 years ago
11

Which best explains whether a triangle with side lengths 5 cm, 13 cm, and 12 cm is a right triangle?

Mathematics
1 answer:
Anon25 [30]3 years ago
6 0
The pythagoream theorem (not sure if I spelled that right...lol): a^2 + b^2 = c^2......where c is the hypotenuse and it is the longest side

5^2 + 12^2 = 13^2
25 + 144 = 169
169 = 169....(correct)...so it is a right triangle
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What is the volume of the rectangular prism?<br> 5 cm<br> 8 cm<br> 1 cm<br> cm3
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Step-by-step explanation:

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3 years ago
Todd wants to make a snack of a number of grapes and slices of cheese. He knows that each grape has 2 calories. The slice of che
Viefleur [7K]

Answer:

The answer to your question is:

a) C = 2g + 155c

b) g = 25 grapes

c) c = 3

Step-by-step explanation:

Data

grapes = g = 2 calories

cheese = c = 155 calories

a) Equation, we consider the amount of grapes and the calores given.

Total calories = C = 2g + 155c

b) We consider that the slices of cheese stays the same

                       2g + 155 = 205

                       2g = 205 -155

                        2g = 50

                        g = 50/2 = 25 grapes

c) Then the number of grapes stays the same

                       2(25) + 155c = 515

                       50 + 155c = 515

                       155c = 515 - 50

                        155c = 465

                        c = 465/155

                        c = 3 slices of cheese

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Lines A and B are parallel to each other. Lines B and E are perpendicular to each other. 
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Find the nature of the root
GREYUIT [131]

Answer:

1) Real and same.

2) Real and distinct

3) Real and distinct

4) Real and distinct

5) Real and distinct

6) Real and distinct

7) Real and distinct

8) Real and distinct

Step-by-step explanation:

If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.

If D > 0, then there will be two different real roots.

If D = 0, then two same and real roots.

If D < 0, then two distinct but imaginary roots.

Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.

2) 5x² - x = 4x² + 2

⇒ x² - x - 2 = 0

It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.

3) \frac{2}{x} + \frac{3}{x} = x - 4

⇒ 5 = x² - 4x

⇒ x² - 4x - 5 = 0.

So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.

4) x(x - 5) = 4(5 - x)

⇒ x² - x - 20 = 0

Hence, D = (-1)² - 4 × 1 × (-20) = 81

So, the roots will be real and distinct.

5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45

So, the roots will be real and distinct.

6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.

Hence, the roots will be real and distinct.

7) 5x² - 6 = 13x

⇒ 5x² - 13x - 6 = 0

So, D = (-13)² - 4 × 5 × (-6) = 289

So, the roots will be real and distinct.

8) x² - x = 3(x + 7)

⇒ x² - 4x - 21 = 0

It has D = (-4)² - 4 × 1 × (-21) = 100

So, the roots will be real and distinct. (Answer)

6 0
3 years ago
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