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Andru [333]
3 years ago
14

Rammy has $9.60 to spend on some peaches and a gallon of milk. Peaches

Mathematics
1 answer:
sergeinik [125]3 years ago
5 0

Answer:

\large \boxed{\text{5.00 lb}}

Step-by-step explanation:

\begin{array}{rcl}1.20x + 3.60 & \leq & 9.60\\1.20x & \leq & 6.00\\x &\leq & \mathbf{5.00}\\\end{array}\\\text{Rammy can buy $\large \boxed{\textbf{5.00 lb}}$ of peaches.}

Check:

\begin{array}{rcl}1.20(5.00) + 3.60 & \leq & 9.60\\6.00 + 3.60 & \leq & 9.60\\9.60 & \leq & 9.60\\\end{array}

OK.

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Which of the following expressions represents the verbal description below?
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((4x-6)^3)/(2(x+1))

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Which store charges more per towel? P=35 + 18n
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The man rushes through your door with his fists in the air, all of the sudden your vision went black, and he had stolen your points.

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3 years ago
How can you divide 360 into 6 parts?
tankabanditka [31]
Use long division if you have to.
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5 0
3 years ago
Read 2 more answers
X^3 - y^3 = 218 where x and y are positive integers. Find x+y
finlep [7]

The value of x + y is 12

<h3>How to solve for x + y?</h3>

The equation is given as:

x^3 - y^3 = 218

Make x^3 the subject

x^3 = 218 + y^3

The next cubic expression greater than 218 is 343.

So, we assume

x^3 = 343

This gives

x = 7

Substitute x^3 = 343 in x^3 = 218 + y^3

343 = 218 + y^3

Evaluate the like terms

y^3 = 125

This gives

y = 5

So, we have:

x + y = 7 + 5

Evaluate

x + y = 12

Hence, the value of x + y is 12

Read more about equations at:

brainly.com/question/2972832

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8 0
2 years ago
Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]
Stels [109]

Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

7 0
3 years ago
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