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r-ruslan [8.4K]

2 years ago

14

A post it note has an area of A square inches. The width of the post it note is 4 inches.which equation represents X the length

of the post it note in inches A X=4/A. B. X=A+2(4) C. X=A/4. D. X=A-2(4)

1 answer:

SVEN [57.7K]2 years ago

6 0

Answer:

Option C. X=A/4

Step-by-step explanation:

we know that

The area of the rectangular post it note is equal to

$A=LW$

where

L is the length

W is the width

In this problem we have

$W=4\ in$

$L=X\ in$

substitute

$A=4X$

solve for x

That means ----> isolate the variable x

Divide by 4 both sides

$X=\frac{A}{4}$

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antiseptic1488 [7]

Answer:

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Step-by-step explanation:

Given: 5^3– 10^2 = X(8 – 2) + 2X – 3X

Solving the equation to find the value of x.

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Using the distributive property of multiplication, multiplying x with 8 and -2

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adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = $360^{o}$ - $324^{o}$

= 0 $36^{o}$

Sum of angles in a triangle = $180^{o}$

<P + <Q + <S = $180^{o}$

048° + 0 $36^{o}$ + <S = $180^{o}$

$84^{o}$ + <S = $180^{o}$

<S = $180^{o}$ - $84^{o}$

= $96^{o}$

<S = $96^{o}$

Applying the Sine rule,

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$\frac{y}{Sin P}$ = $\frac{z}{Sin S}$

$\frac{y}{Sin 48^{o} }$ = $\frac{17}{Sin 96^{o} }$

$\frac{y}{0.74314}$ = $\frac{17}{0.99452}$

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⇒ x = $\frac{9.992430}{0.99452}$

= 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

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