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Assoli18 [71]
2 years ago
9

BRAINLIEST AND 40 POINTS PLZ HELP I NEVER GET HELP

Mathematics
2 answers:
mezya [45]2 years ago
6 0
6.40 units

You use the squaring method to fill in the triangle.

Count the sides 4 up and 5 across
Square them (16 and 25)
Add them together 16+25=41
Find the square root of 41 (6.40)

Good luck!
iren2701 [21]2 years ago
3 0

Answer:

6.4

Step-by-step explanation:

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Maggie's brother is three years younger than twice her age. The sum of their ages is 24 how old is maggie?
Vanyuwa [196]

The variable for Maggie will be M and 2x-3 be her younger brothers age (<u>Twice her age). </u>We then would turn this into <u>an algebraic problem</u>. (m+2x -3=24).

3x-3=24, we would add 3 to both sides (-3 and 24). 24 + 3 equals 27, we now have to divide -3x and 27 on both sides, which equals <u>9. (x=9)</u>



6 0
3 years ago
Michelle buys a motorcycle for $10,000. The motorcycle loses 20% of its value each year. The depreciation of the motorcycle is s
makkiz [27]
You can use cross product to slove this equation.

3 0
3 years ago
Marty made a $220 bank deposit using $10 bills and $5 bills. She gave the teller a total of 38 bills, how many $5 bills were in
Luda [366]

ANSWER: 32 five-dollar bills

======

EXPLANATION:

Let x be number of $5 bills

Let y be number of $10 bills

Since we have total of 38 bills, we must have the sum of x and y be 38

x + y = 38 (I)

Since the total amount deposited is $220, we must have the sum of 5x and 10y be 220 (x and y are just the "number of" their respective bills, so we multiply them by their value to get the total value):

5x + 10y = 220 (II)

System of equations:

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ 5x + 10y &= 220 && \text{(II)} \end{aligned} \right.

Divide both sides of equation (II) by 5 so our numbers become smaller

\left\{ \begin{aligned} x + y &= 38 && \text{(I)} \\ x + 2y &= 44 && \text{(II)} \end{aligned} \right.

Rearrange (I) to solve for y so that we can substitute into (II)

\begin{aligned} x + y &= 38 && \text{(I)} \\ y &= 38 - x \end{aligned}

Substituting this into equation (II) for the y:

\begin{aligned} x + 2y &= 44 && \text{(II)} \\ x + 2(38 - x) &= 44\\ x + 76 - 2x &= 44 \\ -x &= -32 \\ x &= 32 \end{aligned}

We have 32 five-dollar bills

======

If we want to finish off the question, use y = 38 - x to figure out number of $10 bills

y = 38 - 32 = 6

32 five-dollar bills and 6 ten-dollar bills

7 0
3 years ago
Find the probability of rolling a twelve first and then a three when a pair of dice is rolled twice?
yawa3891 [41]
Total number of possible outcomes = 36

Total number of outcomes with a sum of 12 = 1
{6,6)

Total number of outcomes with a sum of 3 = 2
{1,2}{2,1}

P(12 first, then 3) = (1/36)(2/36) = 1/648

Answer: The probability is 1/648
6 0
3 years ago
Read 2 more answers
A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)= at² + bt + c, where t is the
frutty [35]

Answer:

<h2>a) a = -3, b = 18, c = 48; </h2><h2>s(t) = -3t²+18t+48</h2><h2>b) 48m</h2><h2>c) 8secs</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

'A rock was thrown from the top of a cliff such that its distance above sea level was given by s(t)=at²+bt+c, where t is the time in seconds after the rock was released. After 1 second the rock was 63 m above sea level, after 2 seconds 72 m, and after 7 seconds 27 m. a. Find a, b and c and hence an expression for s(t). b. Find the height of the cliff. c. Find the time taken for the rock to reach sea level.'

Given the equation of the distance modelled as s(t)=at²+bt+c

If after 1 second the rock was 63 m, then 63 = a+b+c

If after 2 seconds, the distance was 72 m then 72 = 4a+2b+c

Also if after 7 seconds, the distance is 27 m, then 27 = 49a+7b+c

i) Solving the 3 equations simultaneously to get a, b and c we have;

a+b+c = 63 ... 1

4a+2b+c = 72 ...2

49a+7b+c = 27 ...3

Subtracting 2 from 1 and 3 from 2 we will generate 2 new equations as shown;

eqn 2- eqn1: 3a + b = 9...4

eqn 3- eqn 2: 45a + 5b = -45

eqn 3- eqn 2: 9a+b = -9 ... 5

solving 4 and 5 simultaneously

3a + b = 9 ...4

9a+b = -9 ... 5  

Taking the difference of 4 and 5 we have

6a - -9-9

6a = -18

a = -3

substituting a = -3 into equation 4 to get b we have;

3(-3)+b = 9

-9 + b = 9

b = 9+9

b = 18

substituting a = -3 and b = 18 into equation 1 to get c we have;

-3+18+c = 63

15+c = 63

c = 48

a = -3, b= 18 and c = 48

The distance function will be s(t) = -3t²+18t+48

ii) If the height of the cliff is modelled by the equation  s(t)=at²+bt+c

The height of the cliff is at when t = 0

s(0) = -3(0)²+18(0)+48

s(0) = 48m

The height of the cliff is 48m

iii) At the sea level, the height of the rock will be 0m, substituting this into the modeled equation for the height to get the time we have;

s(t)=at²+bt+c

0 = -3t²+18t+48

3t²-18t-48 = 0

t² - 6t - 16 =0

t² - 8t+2t - 16 = 0

t(t-8)+2(t-8) = 0

(t+2)(t-8) = 0

t = -2 or 8

Taking the positive value of the time, t = 8secs

Time taken for the rock to reach sea level is 8secs

7 0
3 years ago
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