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3 years ago

A pencil and a ruler cost $1.50 together. The ruler costs $1.00 more than the pencil. How much does the pencil cost?

1 answer:

7 0

So we want to know how much does the pencil costs if the ruler costs 1$ more than the pencil and together they cost 1.5$. So lets start with converting dollars into cents. Lets turn this into an equation. Ruler is x and the pencil is y. So together ruler and pencil cost 150cents or: x+y=150 cents. The ruler costs 100 cents more than the pencil, or: (x+100 cents ) + y = 150 cents. So we see that we need to put the values of x=25 cents and y=25 cents to get the ruler to cost 125 cents which is 100 cents or 1$ more than the pencil. So the pencil costs 25 cents.

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a shopkeeper sold a pen at loss of 20%. if he had sold it for rs 10 more, he would have gain 5%.find the cost price of pen

AlexFokin [52]

Rs 200
since the questions states that if he sold it for 10 more, he would have gained 5%, 10=5%
1%=2
100%=200

8 0

3 years ago

The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r

mash [69]

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$

3 0

2 years ago

Help me quick i need helppp

uysha [10]

$\\ \sf\longmapsto (m-8)+(m-8)+(m-8)+(m-8)+(m-8)+(m-8)=12$

$\\ \sf\longmapsto 6(m-8)=12$

$\\ \sf\longmapsto 6m-48=12$

$\\ \sf\longmapsto 6m=12+48$

$\\ \sf\longmapsto 6m=60$

$\\ \sf\longmapsto m=\dfrac{60}{6}$

$\\ \sf\longmapsto m=10$

5 0

3 years ago

Read 2 more answers

Please answer #5 A-C!

Cloud [144]

Answer:a

Step-by-step explanation:I know

4 0

3 years ago

Simplify 4c2+6c-3c2-2c-3

NikAS [45]

Let's simplify step-by-step.4c2+6c−3c2−2c−3=4c2+6c+−3c2+−2c+−3Combine Like Terms:=4c2+6c+−3c2+−2c+−3=(4c2+−3c2)+(6c+−2c)+(−3)=c2+4c+−3

Answer: =c2+4c−3

6 0

3 years ago