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pentagon [3]
3 years ago
8

Which statement about a trapezoid is always true?

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
4 0
The answer is <span>It has at least one pair of parallel sides. </span>
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Prove that the value of the expression: (125^2+25^2)(5^2–1) is divisible by 39
prisoha [69]

Answer:

(1252+252)(52−1)

=(1252+252)(52+−1)

=(1252)(52)+(1252)(−1)+(252)(52)+(252)(−1)

=390625−15625+15625−625

=390000

divide by 39 is

=10000

4 0
3 years ago
Simplify.
Mekhanik [1.2K]
6a - (b - (3a - (2b + c + 4a - (a + 2b - c))))
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6a - (b - (3a - (2b - 2b + 4a - a + c + c)))
6a - (b - (3a - (3a + 2c)))
6a - (b - (3a - 3a - 3c))
6a - (b - 3a + 3a + 3c)
6a - (b + 3c)
6a - b - 3c

x³ + x² - 25x - 25
x²(x) + x²(1) - 25(x) - 25(1)
x²(x + 1) - 25(x + 1)
(x² - 25)(x + 1)
(x² - 5x + 5x - 25)(x + 1)
(x(x) - x(5) + 5(x) - 5(5))(x + 1)
(x(x - 5) + 5(x - 5))(x + 1)
(x + 5)(x - 5)(x + 1)

36x² + 60x + 25
36x² + 30x + 30x + 25
6x(6x) + 6x(5) + 5(6x) + 5(5)
6x(6x + 5) + 5(6x + 5)
(6x + 5)(6x + 5)
(6x + 5)²
8 0
3 years ago
Read 2 more answers
ASAP please...........​
Veseljchak [2.6K]

Answer:

I think it would be b

Step-by-step explanation:

He doesn't want to spend more than 50 and he needs 8 for lunch so the 8 dollars and the book can't go over 50.

Its either a or b im just not sure if its plus 8 or minus 8. sorry

4 0
3 years ago
Read 2 more answers
Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).
fomenos

Answer:

\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is 2a=|4--10|.

\implies 2a=|14|

\implies 2a=14

\implies a=7

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is (\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)

We need to use the relation a^2+b^2=c^2 to find b^2.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

c=|6--3|=9

\implies 7^2+b^2=9^2

\implies b^2=9^2-7^2

\implies b^2=81-49

\implies b^2=32

We substitute these values into the standard equation of the hyperbola to obtain:

\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1

\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1

7 0
3 years ago
¿cuál es el vértice de una parábola que corta los ejes en los puntos (-1,0) , (5,0), (0,-10) ?​
kvasek [131]

Answer:

vertex: (2,-18)

Step-by-step explanation:

y = ax² + bx + c

(-1,0) : a - b + c = 0       ...(1)

(5,0) : 25a +5b +c = 0  ...(2)

(0,-10): 0a + 0b + c = -10        c=-10

(1) x5: 5a - 5b + 5c = 0  ...(3)

(2)+(3): 30a + 6c = 0          30a = -6c = 60        a = 2

(1): 2 - b -10 = 0     b = -8

Equation: y = 2x² - 8x -10 = 2 (x² -4x +4) - 18 = 2(x-2)² -18

equation: y = a(x-h)²+k     (h,k): vertex

vertex: (2,-18)

3 0
3 years ago
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