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kvv77 [185]
2 years ago
13

HELP DUE IN 20 MINS!!!!!!

Mathematics
1 answer:
Amiraneli [1.4K]2 years ago
5 0

Answer:

10 miles

Step-by-step explanation:

  1. Set up a proportion: \frac{15}{1.5} = \frac{x}{1}  
  2. Cross multiply, then divide: 15 × 1 = 15, 15 ÷ 1.5 = 10
  3. So, Jessica can run 10 miles in 1 hour

I hope this helps!

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Find the minimum value of the <br> parabola<br> y = x2 − 4x − 5 .
enot [183]

Answer:

-2x-5

Step-by-step explanation:

8 0
3 years ago
Mr. Penn wants to buy pencils and erasers for the students in his class. If pencils come in packs of 20 and erasers in packs of
joja [24]
5 packs of erasers and 3 packs of pencils each will have 60
4 0
3 years ago
Which constant could each equation be multiplied by to eliminate the x variable in this system of equations? 4x + 5y = 62 5x − 3
qwelly [4]

Answer:

Multiply equation 4x+5y=62 by 5 and 5x-3y=22 by 4

Step-by-step explanation:

Given:

Equations are 4x+5y=62,5x-3y=22

To find: constant by which each equation should be multiplied to eliminate the variable x in the given system of equations

Solution:

4x+5y=62...(i)\\5x-3y=22...(ii)

Multiply equation (i) by 5 and (ii) by 4 to get the following equations:

20x+25y=310\\20x-12y=88

On subtracting these equations, we get

(20x+25y)-(20x-12y)=310-88\\\\13y=222

So, the variable x gets eliminated

7 0
3 years ago
When two distinct circles lie in the same plane, what is the largest number of points at which they can intersect?
Rama09 [41]
All the points if they lay on top of eachother

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6 0
3 years ago
A rectangular box without a lid is to be made from 48 m2 of cardboard. Find the maximum volume of such a box. SOLUTION We let x,
tatiyna

Answer:

The maximum volume of such box is 32m^3

V = x×y×z = 32 m^3

Step-by-step explanation:

Given;

Total surface area S = 48m^2

Volume of a rectangular box V = length×width×height

V = xyz ......1

Total surface area of a rectangular box without a lid is

S = xy + 2xz + 2yz = 48 .....2

To be able to maximize the volume, we need to reduce the number of variables.

Let assume the rectangular box has a square base,that means; length = width

x = y

Substituting y with x in equation 1 and 2;

V = x^2(z) ....3

x^2 + 4xz = 48 .....4

Making z the subject of formula in equation 4

4xz = 48 - x^2

z = (48 - x^2)/4x .......5

To be able to maximize V, we need to reduce the number of variables to 1, by substituting equation 5 into equation 3

V = x^2 × (48 - x^2)/4x

V = (48x - x^3)/4

differentiating V with respect to x;

V' = (48 - 3x^2)/4

At the maximum point V' = 0

V' = (48 - 3x^2)/4 = 0

Solving for x;

3x^2 = 48

x = √(48/3)

x = √(16)

x = 4

Since x = y

y = 4

From equation 5;

z = (48 - x^2)/4x

z = (48 - 4^2)/4(4)

z = 32/16

z = 2

The maximum volume can be derived by substituting x,y,z into equation 1;

V = xyz = 4×4×2 = 32 m^3

7 0
3 years ago
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