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murzikaleks [220]
3 years ago
5

At a jewelry store, the price p of a gold necklace varies directly with its length l. Also, the weight w of a necklace varies di

rectly with its length. Show that the price of a necklace varies directly with its weight.
Mathematics
1 answer:
slega [8]3 years ago
4 0

Since p varies directly with l,

p = kl     (1)

Also, since w varies directly with l,

w = ml  

or, l = w/m

Substitute in (1), we get,

p = k (w/m)

= (k/m) w

So, the price varies directly with weight.

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What is the solution to −1/4(d+1)<2
kogti [31]

Answer:

d > -9

Step-by-step explanation:

-\dfrac{1}{4}(d + 1) < 2

First, multiply both sides by -4. You must also change the inequality sign since you are multiplying both sides by a negative number.

-4 \times \left(-\dfrac{1}{4}\right)(d + 1) > 2 \times (-4)

d + 1 > -8

Subtract 1 from both sides.

d + 1 - 1 > -8 - 1

d > -9

5 0
2 years ago
Yall please help meee
madam [21]

Answer-

x=0;y=3

x=1;y=-1

x=2;y=-5

x=3;y=-9

7 0
3 years ago
Add 3/4+(−2 1/2) using the number line.
Neko [114]

Answer:

-1 3/4

Step-by-step explanation:

6 0
3 years ago
(4x + 4)(ax – 1) – x2 +4
Oksana_A [137]

Answer:

b=-3

Step-by-step explanation:

If the expression simplifies to bx that means the x^{2} terms and the constant terms must be cancel out.

Simplify it first.

(4x+4)(ax-1)-x^{2} +4\\=(4ax^{2} -4x+4ax-4)-x^{2} +4\\=4ax^{2} -x^{2} -4x+4ax-4+4\\=4ax^{2} -x^{2} -4x+4ax

We know –4 + 4 will cancel out. If we simplify this expression to only an x term, then the x^{2} terms should be cancelled. Therefore, we say that 4ax^2 – x^2 = 0.

4ax^{2} -x^{2} =0\\x^{2} (4a-1)=0\\4a-1=0\\4a=1\\a=\frac{1}{4}

If we put a = ¼, then we can find the value of b:

=4(\frac{1}{4} )x^{2} -x^{2} -4x+4(\frac{1}{4} )x\\=x^{2} -x^{2} -4x+x\\ (cancel out x^{2} terms)\\

=-3x

bx=-3x\\      if the  expression is equivalent to bx

Therefore, b = –3.

8 0
3 years ago
What is the solution to -2(8×-4)&lt;2x+5
Monica [59]
Distribute the -2 to get 

-16x + 8 < 2x + 5

3 < 18x 

1/6 < x

so x > 1/6
7 0
3 years ago
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