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3 years ago

PLSSS HELP!! ASAP! GIVING BRAINLIST

Mathematics

1 answer:

EastWind [94]3 years ago

6 0

$\sqrt{180}$

$\sqrt{ {6}^{2} \times 5 }$

$\sqrt{ {6}^{2} } \sqrt{5}$

$6 \sqrt{5}$

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Robbie was given $80 to spend at the Fair. His admission to the park costs $15.50 and each ride cost $5. He anticipates the cost

Ad libitum [116K]

Answer:

7 or 8

Step-by-step explanation:

Robbie was given $80.

His admission costs $15.50

80 - 15.50 = $64.50

He anticipates the cost of food at the fair will be $25.

64.50 - 25 = $39.50

What is the maximum number of rides he can take at the Fair?

(Total $ after admission and cost of food/ ride cost)

$39.50 / $5 = 7.9

4 0

3 years ago

Read 2 more answers

0.12x - 1.5= 0.24x - 0.06

Sophie [7]

To solve this equation, lets move all of the variables to one side of the equation and the constants to the other.

0.12x - 1.5 = 0.24x - 0.06

Subtract 0.12x from both sides of the equation

-1.5 = 0.12x - 0.06

Add 0.06 to both sides of the equation

-1.44 = 0.12x

Divide both sides by 0.12 to isolate x

x = -12

5 0

3 years ago

Suppose that the position of one particle at time is given by x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π and the position of a second p

Mashcka [7]

Answer:

there is no collision between the particles

Step-by-step explanation:

for the first particle

x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π

for the second particle

x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π

then for the collision

x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t

y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t = 2 cos t →(1/3) = 2

since 1/3 ≠ 2 there is no collision between the particles

6 0

3 years ago

A group of winning ticket holders share equally 90,000,000 lottery. Before the money is divided, three more winning ticket holde

Kruka [31]

Let's assume

people were in the original group of winners as x

A group of winning ticket holders share equally 90,000,000 lottery

Before the money is divided, three more winning ticket holders are declared

so, total number of winning tickes =x+3

each persons share is reduced to 5,000,000

so, total money =5,000,000*(x+3)

and this must be equal to total money

$5000000(x+3)=90,000,000$

now, we can solve for x

and we get

$x+3=18$

Subtrac both sides by 3

$x+3-3=18-3$

$x=15$

so,

number of people were in the original group of winners is 15 ...........Answer

6 0

3 years ago

Q5: The probability that event A occurs is 5/7 and the probability that event B occurs is 2/3 . If A and B are independent event

bekas [8.4K]

Answer:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

Step-by-step explanation:

We define two events for this case A and B. And we know the probability for each individual event given by the problem:

$p(A) = \frac{5}{7}$

$p(B) = \frac{2}{3}$

And we want to find the probability that A and B both occurs if A and B are independent events, who menas the following conditions:

$P(A|B) = P(A)$

$P(B|A) = P(B)$

And for this special case we want to find this probability:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

7 0

3 years ago