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algol [13]
3 years ago
10

Select each unit fraction. A. 6 6 B. 1 8 C. 2 4 D. 1 3 E. 3 4

Mathematics
1 answer:
Drupady [299]3 years ago
3 0
No unit fractions are shown. (Sometimes editing is required when you copy and paste your question text.)

Selections B and D have 1 as their first digit. We suspect those are supposed to be the unit fractions 1/8 and 1/3, respectively.
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What is the distance between the points?! round to the nearest tenth i need to show work
Maru [420]

Answer:

c=\sqrt{52}

Step-by-step explanation:

a=height=6

b=length=4

using pythagorean theorem

a^2+b^2=c^2

36+16=c^2

c=\sqrt{36+16}

c=\sqrt{52}

6 0
2 years ago
Can you please help me with this
erastova [34]
Well there are no specific numbers, but the area of a rectangle is the length x the width, and the width is the area ÷ the length.
7 0
3 years ago
X + 3/2= 2 1/4 whats the answer
kicyunya [14]

Move all terms not containing x to the right side of the equation.

Exact Form:

x=15/4

Decimal Form:

x=3.75

Mixed Number Form:

x=3 3/4

hope this helps

5 0
1 year ago
When 5 randomly selected people all have the same birth month of March.
Doss [256]

Answer:

I know 4 people that are all really close friends and all have birthdays in march.

Step-by-step explanation:

3 0
3 years ago
Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.
il63 [147K]

Answer: The length is 13cm and the width is 7cm

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = W*L

In this case we have:

L = (2*x + 9) cm

W=(3*x + 1) cm

Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

A = (6*x^2 + 2*x + 27*x + 9) cm^2

A = (6*x^2 + 29*x + 9) cm^2

now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm)  = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 =  (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)}  }{2*4} = \frac{-29 \pm 45}{8}

Then the two solutions are:

x = (-29 - 45)/8 = -9.25  (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).

The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

Width = (3*2 + 1)cm = 7cm

4 0
2 years ago
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