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2 years ago

Which of the following numbers are less than nine over four

Mathematics

1 answer:

allochka39001 [22]2 years ago

5 0

Answer:

option B and C

Step-by-step explanation:

option B and C

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How many ducks can fit in a duck pond if there are 24 ducks and 10 lakes

kiruha [24]

You can fit 2 in each or 3 can have some in it

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3 years ago

A bicyclist goes 43.54 miles in 3.5 hours. What is her speed?

Studentka2010 [4]

Divide 43.54 by 3.5
She was going 12.33 miles an hour

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3 years ago

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postnew [5]

Answer:

Area = 168m^2

Step-by-step explanation:

Area of triangle = 1/2×base×height

Area = 1/2bh

= 1/2(8+16)(14)

= 168m^2

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2 years ago

40 POINTS AND BRAINLIEST

ladessa [460]

Answer:

120

Step-by-step explanation:

As x goes from 1 to 2, y increases by 4 from 4 to 8

As x goes from 2 to 3, y increases by 4 from 8 to 12

We increase by 4 each time

The formula for an arithmetic sequence is

an = a1+d(n-1)

where a1 is the first term which is 4

d is the common difference or the number we add each time which is 4

n is the term we are looking for which is 30

a30 = 4 +4 (30-1)

a30 = 4+4(29)

= 4+116

= 120

5 0

3 years ago

Read 2 more answers

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

3 years ago