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Gemiola [76]
1 year ago
15

Hanna shops for socks that cost $2.99 for each pair and blouses that cost $12.99 each. Let x represent the number of pairs of so

cks purchased, and let y represent the number of blouses purchased. Which equation models the purchases she made with $43.92?
x + y = 15.98
x + y = 43.92
43.92x - 2.99y = 12.99
2.99x + 12.99y = 43.92
Mathematics
1 answer:
rusak2 [61]1 year ago
7 0

Answer:

2.99x + 12.99y = 43.92

Step-by-step explanation:

Because, 2.99 is for each pair of socks and she bought x number of that. Same for 12.99 (Y).

In conclusion her total was 43.92 and she could have bought many socks as she'd like and blouses.

These calculations are 100% correct! :)

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In the figure, the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD is . The area of polygon ABCDE is s
Gekata [30.6K]

Step 1

Find the perimeter of rectangle ABDE

we know that

the perimeter of rectangle is equal to

P=2b+2h

In this problem

b=ED=2\ units

h=AE=6\ units

substitute

P=2*2+2*6=16\ units  

Step 2

Find the perimeter of triangle BCD

we know that

the perimeter of triangle is equal to

P=BD+DC+BC

In this problem we have

BD=AE=6\ units

DC=BC

Applying the Pythagoras theorem

DC^{2}=4^{2}+3^{2}

DC^{2}=25

DC=5\ units

substitute

P=6+5+5=16\ units

Find the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD

we have

the perimeter of rectangle is equal to

P=16\ units  

the perimeter of the triangle is

P=16\ units  

so

the ratio is equal to

\frac{16}{16} =1

therefore

<u>the answer Part 1) is the option B</u>

1

Step 3

Find the area of polygon ABCDE

we know that

The area of polygon is equal to the sum of the area of rectangle plus the area of triangle

Area of rectangle is equal to

A=AE*BD=6*2=12\ units^{2}

Area of the triangle is equal to

A=\frac{1}{2}AEh

the height h of the triangle is equal to 4\ units

substitute

A=\frac{1}{2}(6)(4)=12\ units^{2}

The area of polygon is

12\ units^{2}+12\ units^{2}=24\ units^{2}

therefore

<u>the answer part 2) is the option C</u>

24\ units^{2}


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2 years ago
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Belleview high school has 12 chaperones for 198 students on a field trip which of the following schools has the same ratio of ch
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16.5 has the same ratio
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1 year ago
A sequence is defined by mc024-1.jpg, and mc024-2.jpg. What is the seventh term?
abruzzese [7]
Given data :
a₃ = 9/16
aₓ = -3/4 · aₓ₋₁
Where x is the number of terms ('x' is also written as 'n')
To find the 7th term (a₇):

We know that aₓ = -3/4 · aₓ₋₁
So,
a₃ = -3/4 · a₃₋₁
a₃ = -3/4 · a₂
9/16 = -3/4 · a₂
a₂ = 9/16 × -4/3
a₂ = -36/48
a₂ = -3/4

Again,
aₓ = -3/4 · aₓ₋₁
a₄ = -3/4 · a₄₋₁
a₄ = -3/4 · a₃
a₄ = -3/4 · 9/16
a₄ = -27/64
a₄ = -27/64

For a₅,
aₓ = -3/4 · aₓ₋₁
a₅ = -3/4 · a₅₋₁
a₅ = -3/4 · a₄
a₅ = -3/4 × -27/64
a₅ = 81/256

For a₆,
aₓ = -3/4 · aₓ₋₁
a₆ = -3/4 · a₆₋₁
a₆ = -3/4 · a₅
a₆ = -3/4 × 81/256
a₆ = -243/1024

For a₇,
aₓ = -3/4 · aₓ₋₁
a₇ = -3/4 · a₇₋₁
a₇ = -3/4 · a₆
a₇ = -3/4 × -243/1024
a₇ = 729/4096
3 0
2 years ago
Read 2 more answers
Use complete sentences to describe the range of the sine function.
MA_775_DIABLO [31]
The Range of a function is the set of all values that that function can take.

Given the sine function f(x)=sinx,

This function is the function which calculates the sine of the values of x.

According to the definition of the sine of an angle x in the unit circle, 

-1 \leq sinx \leq 1,

so the sine of an angle is always larger or equal to -1, and smaller or equal to 1.

This means that the values that the sine function takes are any values between -1 and 1, inclusive.

This determines the Range of the sine function. 

So the Range of the sine function is [-1, 1]
3 0
2 years ago
In high-school 135 freshmen were interviewed.
timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0
1 year ago
Read 2 more answers
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