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Lesechka [4]
2 years ago
8

Elina and Gustavo leave Cayley H.S. at 3:00 p.m. Elina runs north at a constant speed

Mathematics
1 answer:
Andrej [43]2 years ago
8 0

Answer:

the answer to your question is 'b'

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Solve for x: 1/2 (10x+6)=-2x+10
pickupchik [31]

Answer: x = 1

Step-by-step explanation:

substitute 1 for x

1/2(10(1)+6) = -2(1)+10

1/2(10+6) = -2+10

5+3 = -2+10

8=8

8 0
2 years ago
Mr Thompson uses a brace to support a Gate write and solve an equation to find the measure of the unknown angle.
algol [13]
Use the pythagorean theorem (A squared + b squared = c squared
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3 years ago
En el experimento Extraer una carta de una baraja española se consideran estos sucesos: A= 《Salir as 》 B= 《salir bastos 》 Calcul
Ganezh [65]

Answer:

They are compatible

Step-by-step explanation:

The first thing is to say that an "ace" and that it is a "coarse"

"ace" is card number 1. Group A

"coarse" is a type of the deck, found from number 1 to card 13. Group B

Thus:

 Calculate A U B:

1 to 13 + 1 of the other types of cards in the deck.

At intersection B:

1 of "coarse"

Therefore, if group A is compatible with group B

7 0
3 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
(7 x 10^−12)(4 x 10^5) in scientific notation
Radda [10]
<span>(7 x 10^−12)(4 x 10^5)
= (7x4) x 10^-7
= 28 x 10^-7
= 2.8 x 10^-6</span>
3 0
3 years ago
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