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mash [69]
3 years ago
6

What is a object accelerated if its moving in 30m and stops at 5 seconds

Mathematics
1 answer:
Ivan3 years ago
8 0
6 meter / second square.

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Please hurry I will give brainless answer if correct
Natali [406]

Answer:

we have to be able to read it first tho. we can't see what the problem is so we need to see it please and thank you...

7 0
2 years ago
In a bag of candy, Marsha found that there were 7 red, 8 blue, 5 yellow, and 9 greens pieces of candy. Marsha put all of the pie
son4ous [18]

Answer:

Step-by-step explanation:

1) find the total candies

7 + 8 + 5 + 9 = 29

2) a probability can be expressed with a fraction whose denominator is the total candies, while the numerator represents the candies that we want to find

- red = 7/29

- blue = 8/29

- yellow = 5/29

- green = 9/29

3) compound probability

(red + blue)/29 = 15/29

(yellow + green)/29 = 14/29

(blue + yellow + green)/29 = 22/29

7 0
2 years ago
I need help with this question i dont understand it
RUDIKE [14]
The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2}  }
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
\sqrt{(3-2) ^{2}+(4-3) ^{2}  }
which simplifies to
\sqrt{(1 )^{2}+(1) ^{2}  } which is \sqrt{2}
So that is the length of both MN and LM.  So far our perimeter is \sqrt{2} + \sqrt{2}=2 \sqrt{2}
Now let's use the same formula to find out the length of one of the longer segments:
\sqrt{(5-3) ^{2} +(3-2) ^{2} }
which simplifies down to
\sqrt{(2) ^{2} +(1) ^{2} }
which is of course \sqrt{5}
Since we have 2 of those lengths, \sqrt{5} + \sqrt{5}=2 \sqrt{5}
So our perimeter is, in the end, 2 \sqrt{2}+2 \sqrt{5}
That's the third choice down
7 0
2 years ago
Caroline has some dimes and some quarters. She has a maximum of 15 coins worth at least $2.85 combined. If Caroline has 3 dimes,
gayaneshka [121]

Answer:

11, 12.

Step-by-step explanation:

Let q represent number of quarters.

We have been given that Caroline has a maximum of 15 coins worth at least $2.85 combined. We are also told that Caroline has 3 dimes. This means that total coins are less than or equal to 15.

We can represent this information in an inequality as:

q+3\leq 15...(1)

We are also told that the coins worth at least $2.85 combined. This means that the worth of all coins is greater than or equal to 2.85.

We know that each dime is worth $0.10 and each quarter is worth $0.25.

0.25q+3(0.10)\geq 2.85...(2)

Now, let us solve our system of inequalities.

From 1st inequality, we will get:

q+3-3\leq 15-3

q\leq 12

From 2nd inequality, we will get:

0.25q+0.30\geq 2.85

0.25q+0.30-0.30\geq 2.85-0.30

0.25q\geq 2.55

\frac{0.25q}{0.25}\geq \frac{2.55}{0.25}

q\geq 10.2

Upon combining our both inequalities, we will get:

10.2\leq q\leq 12

This means that numbers of quarters would be greater than or equal to 10.2 and less than or equal to 12.

Since we cannot have 0.2 of a coin, therefore, Caroline could have 11 or 12 quarters.

6 0
3 years ago
Find AB distance. Please help
Svet_ta [14]

Answer:

Solution,

Let,(x1,y1)=(-4,-3)

     (x2,y2)=(3,5)

Using distance formula,

AB=√(x2-x1)²+(y2-y1)²

    =√(3+4)²+(5+3)²

    =√49+64

    =√113 units.

Step-by-step explanation:

7 0
2 years ago
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