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3 years ago

A jewelry shop sells 240 necklaces in a month.

Mathematics

1 answer:

zhuklara [117]3 years ago

5 0

Answer:3 to 2

Step-by-step explanation:

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Karim wants to buy a shirt that normally sells for $46.50. It is on sale for One-third off. How much does Karim need to pay to b

Volgvan

Answer:

Step-by-step explanation:

First I took the original price of the shirt (46.50) since it is going one third off I divided it by 3 (46.50/3=15.50)

I took away the one third I divided from it (46.50-15.50=31)

I got 31 dollars as my final answer.

Now Karim needs to buy the shirt for $31.

8 0

3 years ago

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4. -2(x+3y)= 18 Plz help me!!!

OLEGan [10]

Answer:

x+3y=-9

Step-by-step explanation:

-2(x+3y)=18

x+3y=18/-2

x+3y=-9

7 0

3 years ago

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A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

Answer:

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

3 years ago

Devin bought 2 candy bars and she wants to share them equally amongst her 7 friends.How much of the cany bar will each friend re

lbvjy [14]

0.28. 2 divide by 7 =

5 0

3 years ago

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I'm stuck and really need help. I can't figure out the rest.

Ad libitum [116K]

Answer:

The next statements are are;

Statement ${}$ Reason

Segment $\overline {DC}$ ≅ Segment $\overline {AB}$ ${}$ CPCTC

$\overline {AD}$ ≅ $\overline {BC}$ and $\overline {DC}$ ≅ $\overline {AB}$ , therefore;

ABCD is a parallelogram ${}$ Parallelogram side theorem

Step-by-step explanation:

Statement ${}$ Reason

Segment $\overline {DC}$ ≅ Segment $\overline {AB}$ by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

Given that $\overline {AD}$ ≅ $\overline {BC}$ and we have that $\overline {DC}$ ≅ $\overline {AB}$ , we get;

ABCD is a parallelogram ${}$ Parallelogram side theorem

The parallelogram side theorem states that a quadrilateral is a parallelogram if it has two pairs or sets of congruent opposite sides.

4 0

3 years ago