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trasher [3.6K]
3 years ago
13

If two similar rectangles has a side ratio of 1:6, and the area of the smaller rectangle is 11cm squared, what is the area of th

e larger rectangle
Mathematics
1 answer:
castortr0y [4]3 years ago
4 0
11 x 17? I would believe
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A piggy bank contains pennies, nickels, and dimes. The number of dimes is 15 more than the number of nickels, and there are 140
amm1812

Answer:

Option D is correct.

There are 34 nickels in the piggy bank.

Step-by-step explanation:

A piggy bank contains pennies, nickels and dimes.

Let the number of pennies be p

Let the number of nickels be n

Let the number of dimes be d

Also, note that 1 penny = $0.01

1 nickel = $0.05

1 dime = $0.10

- The number of dimes is 15 more than the number of nickels.

d = 15 + n

- There are 140 coins altogether totaling $7.17.

p + n + d = 140

0.01p + 0.05n + 0.1d = 7.17

Bringing the 3 equations together

d = 15 + n (eqn 1)

p + n + d = 140 (eqn 2)

0.01p + 0.05n + 0.1d = 7.17 (eqn 3)

Substitute (eqn 1) into (eqn 2)

p + n + d = 140

p + n + (15 + n) = 140

p + 2n + 15 = 140

p = 140 - 15 - 2n = 125 - 2n

p = 125 - 2n (eqn 4)

Substitute (eqn 1) and (eqn 4) into (eqn 3)

0.01p + 0.05n + 0.1d = 7.17

0.01(125 - 2n) + 0.05n + 0.1(15 + n) = 71.7

1.25 - 0.02n + 0.05n + 1.5 + 0.1n = 7.17

0.1n + 0.05n - 0.02n + 1.5 + 1.25 = 7.17

0.13n + 2.75 = 7.17

0.13n = 7.17 - 2.75 = 4.42

0.13n = 4.42

n = (4.42/0.13) = 34

d = 15 + n = 15 + 34 = 49

p = 125 -2n = 125 - (2×34) = 125 - 68 = 57

Hence, there are 57 pennies, 34 nickels and 49 dimes in the piggy bank.

Hope this Helps!!!

7 0
3 years ago
I am confused about how to do this problem and I don't think it should be hard, but I just don't know how to approach it, so if
solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) (0.875, 0.676)

d) (0, 1.235)

Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

In the numerator:

r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8.  To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

5 0
3 years ago
What’s the correct answer for this?
Mice21 [21]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

First, you'd work out the centre value

It would be the opposite value to what is in the brackets

Thus, the centre value is (2, -4)

Plot this value on a graph

In order to find the radius, you must square root the 25

Thus, the radius would be 5

Plot some points with a radius of 5 from the centre and then draw the circle

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

8 0
3 years ago
write a multiplication problem using two decimals that has a product greater than 0.1 but less than 0.2
wariber [46]
So the question is asking to write a multiplication problem that uses two decimals that has a product greater than 0.1 but less than 0.2 and in my further computation, I would give an example of 0.5*1.5 equals 0.75.  I hope you are satisfied with my answer and feel free to ask for more 

6 0
4 years ago
Read 2 more answers
Write 144 as a product of it's prime number. <br>plz faster​
Vlad [161]

Answer:

The prime factors are only

2

and

3

As the product of prime factors:

144

=

2

×

2

×

2

×

2

×

3

×

3

5 0
3 years ago
Read 2 more answers
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