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Wittaler [7]
3 years ago
9

AT LEAST how many cubes like this one need to be joined together to form another LARGER SIZED CUBE?

Mathematics
2 answers:
FrozenT [24]3 years ago
8 0

Answer:

Hi

Step-by-step explanation:

Alex Ar [27]3 years ago
7 0

Answer:

25?

Step-by-step explanation:

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Find the length of QP given Q is the midpoint of XF, PQ =2X+1, XF=7X-4, PF=X
nadezda [96]

Answer:

Therefore the length of QP = 3.4 units

Step-by-step explanation:

Given:

PQ = 2x + 1

XF = 7x - 4

PF = x

Q is the mid poimt of XF

∴ XQ = QF

QF = PQ - PF  ..........( Q - F - P )

     = 2x + 1 - x

∴ QF      = x + 1

∴ XQ = QF = x + 1

TO Find:

QP = ?

Solution:

By Addition Property we have

XP = XQ + QF+PF ..........(X-Q-F-P)\\\\

XF + PF =XQ + QF+PF ..........(X-Q-F-P)\\

Substituting the given values in above equation we get

(7x - 4) + x = (x +1) + (x +1) + x

8x -4 = 3x +2

8x - 3x + 4 + 2

5x = 6

∴ x = \frac{6}{5}

Now we require

QP = (2x + 1)

∴ QP = 2\times \frac{6}{5} +1\\\\QP = \frac{12+5}{5} \\\\QP =\frac{17}{5} \\\\\therefore QP = 3.4\ unit

Therefore the length of QP = 3.4 units

3 0
3 years ago
By selling a radio with 12% profit , the profit
Setler79 [48]

Answer:

the Radio was 52.8%

Step-by-step explanation:

60-12%=52.8

5 0
3 years ago
What is the value of 7 in 6,735?
JulijaS [17]
First, you have to break the number down by expanding it to find the different number values.

          6,735
              ↓
6,000 + 700 + 30 + 5

This shows us the values of the 7 in 6,735 is 700.

Answer: 700

6 0
3 years ago
Express the terms of the following geometric sequence recursively.
BabaBlast [244]

Answer:

The most correct option for the recursive expression of the geometric sequence is;

4. t₁ = 7 and tₙ = 2·tₙ₋₁, for n > 2

Step-by-step explanation:

The general form for the nth term of a geometric sequence, aₙ is given as follows;

aₙ = a₁·r⁽ⁿ⁻¹⁾

Where;

a₁ = The first term

r = The common ratio

n = The number of terms

The given geometric sequence is 7, 14, 28, 56, 112

The common ratio, r = 14/7 = 25/14 = 56/58 = 112/56 = 2

r = 2

Let, 't₁', represent the first term of the geometric sequence

Therefore, the nth term of the geometric sequence is presented as follows;

tₙ = t₁·r⁽ⁿ⁻¹⁾ = t₁·2⁽ⁿ⁻¹⁾

tₙ =  t₁·2⁽ⁿ⁻¹⁾ = 2·t₁2⁽ⁿ⁻²⁾ = 2·tₙ₋₁

∴ tₙ = 2·tₙ₋₁, for n ≥ 2

Therefore, we have;

t₁ = 7 and tₙ = 2·tₙ₋₁, for n ≥ 2.

4 0
3 years ago
Find the area of this figure to the nearest hundredth. Use 3.14 to approximate pi.
Fantom [35]

Answer:

49.12 square units

Step-by-step explanation:

Area of triangle = \dfrac{1}{2}*base *height

                          =\dfrac{1}{2}*6*8=3*8 = 24 \ sq.units

Circle:

d = 8 units

r = 8/2 = 4 units

Area \ of \  semicircle = \dfrac{1}{2}\pi r^{2}\\

                                =\dfrac{1}{2}*3.14*4*4=3.14*2*4= 25.12 \ sq.units

Area of the figure = area of triangle +area of semicircle

   = 24 + 25.12

    = 49.12 sq. units

4 0
3 years ago
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