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meriva
2 years ago
10

Joe gained 5 pounds. John gained 14 pounds. Jordan lost 20 pounds. Josh lost 11 pounds. Who had the biggest change in weight? A.

Jordan B. Joe C. John D. Josh
Mathematics
2 answers:
photoshop1234 [79]2 years ago
8 0

Answer:I think Jordan is this a trick question

Step-by-step explanation:

Marrrta [24]2 years ago
7 0
Jordan has the biggest weight change
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3 years ago
2. Given: HM = VM, and ZH and cV Triangle congruence criteria,
STALIN [3.7K]
Answer: SAS is the correct criteria

Explanation:

Angles VMU and GMH are congruent by the Vertical Angles Theorem. Given that angles UVM and GHM are congruent because they are both right angles, we now have two pairs of corresponding angles. Also given that sides HM and VM are congruent, we now have two corresponding pairs of congruent angles and a pair of congruent sides.Therefore, your best option is the ASA postulate, which states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent. Therefore, we have a corresponding angle, a corresponding side, and another corresponding angle in triangle GHM, which is congruent to its corresponding angle, a corresponding side, and another corresponding angle in triangle UVM.
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3 years ago
Graph the line that contains the point (o,3) and has a slope of 1/2
Katyanochek1 [597]

Answer:

Step-by-step explanation:

y-3=1/2(x-0)=1/2 x

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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}.

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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.

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Zinaida [17]

Answer:

207,460

Step-by-step explanation:

brainiest plz

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3 years ago
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