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3 years ago

Similar circles are always congruent true or false

Mathematics

1 answer:

disa [49]3 years ago

8 0

False. All circles are similar. but not congruent.

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Any help plzzz I’ll appreciate it so much thank u

docker41 [41]

9514 1404 393

Answer:

5√5

Step-by-step explanation:

Use the distance formula:

d = √((x2 -x1)² +(y2 -y1)²)

d = √((-7-(-2))² +(-7-3)²) = √((-5)² +(-10)²) = √(25 +100)

d = √125 = √(25·5)

d = 5√5 . . . . distance between the points

3 0

3 years ago

Simplify -12.5 - (-4.8)

Elan Coil [88]

Answer: -17.5

Step-by-step explanation:Use this rule:

1. -a-b=-(a+b)−a−b=−(a+b).

-(12.5+4.8)−(12.5+4.8)

2.Use the algorithm method.

1 2 . 5

+ 4 . 8

1 7 . 3

3.After simplification, we have:

-(12.5+4.8)=-17.3−(12.5+4.8)=−17.3

4. Therefore, -12.5-4.8=-17.3−12.5−4.8=−17.3.

-17.3−17.3

4 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

Need help with this geometry assignment due soon smart people and mark Brainly

skad [1K]

Answer:

1. a + b + c = 180

2. b + d = 180

3. d = a + c

Step-by-step explanation:

1. a + b + c = 180:

Angles in a triangle add up to 180

2. b + d = 180:

Angles in a straight line equal 180 because they are supplementary angles

3. d = a + c:

ΔACD is an exterior angle, and ΔACB is interior adjacent, hence ∠a and ∠c add up to ∠d

3 0

3 years ago

Read 2 more answers

Please help a friend

choli [55]

Answer:

Step-by-step explanation:

8 0

3 years ago