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aksik [14]
2 years ago
10

The table shows a proportional relationship.

Mathematics
1 answer:
Alenkasestr [34]2 years ago
8 0
(2,2.8)(4,5.6)
slope = (5.6 - 2.8) / (4 - 2) = 2.8/2 = 1.4

y = mx + b
slope(m) = 1.4
use either point (2,2.8)...x = 2 and y = 2.8
now we sub and find b, the y int
2.8 = 1.4(2) + b
2.8 = 2.8 + b
2.8 - 2.8 = b
0 = b

so ur equation is : y = 1.4x + 0..or just y = 1.4x

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A shopkeeper had a profit of ₹ 50 on Monday , a loss of ₹ 20 on Tuesday and of loss of
Oliga [24]

Answer:

12

Step-by-step explanation:

Remark

I can't use a symbol for the currency. I'll just get the answer in numbers. You can add the symbol.

Solution

First day: 50

Second day: - 20

Third day: - 18

50 - 20 - 18 = 12

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Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t
dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: (x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14

<u>Area of the big rectangle</u>: (x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99

The difference is: 11x+85

( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

f(d)=h=-2d^2+7d+4

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

f(0)=h=-2(0)^2+7(0)+4

h=4

The height is 4 meters.

b) 9 meters.

For d=1

f(1)=h=-2(1)^2+7(1)+4

f(1)=h=-2+7+4

h=9

b) The ball travels 4 meters.

But to calculate it, it is when h=0

0=-2d^2+7d+4

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

V=\pi r^2h

V=\pi (2.5)^2 \cdot 5

V=31.25 \pi

V_{c} \approx 98.17 \text{ m}^3

The volume of the sphere:

$V=\frac{4}{3}  \pi r^2$

V_{s} \approx 65.4 \text{ m}^3

The volume of the capsule is approximately 163.57  \text{ m}^3

3 0
3 years ago
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