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3 years ago

3, 12, 48, 192, 768, . . . This sequence has a?

Mathematics

1 answer:

nydimaria [60]3 years ago

8 0

Answer:

common ratio is 4

Step-by-step explanation:

3, 12, 48, 192, 768, . . .

LEts find the common difference

$12-3= 9$

$48-12= 36$

There is no common difference

LEts find out common ratio by dividing the second term by first term

$\frac{12}{3} =4$

$\frac{48}{12} =4$

$\frac{192}{48} =4$

So common ratio is 4

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Which choices are equivalent to the quotient below √16/√8

Darya [45]

Answer:

Step-by-step explanation:

$\frac{\sqrt{16} }{\sqrt{8} } = \sqrt{\frac{16}{8} } = \sqrt{2}$

Hope this helps!

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2 years ago

How many solutions does -46+23=46x+23

Slav-nsk [51]

Answer:

x = -1

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−46+23=46x+23

(−46+23)=46x+23(Combine Like Terms)

−23=46x+23

Step 2: Flip the equation.

46x+23=−23

Step 3: Subtract 23 from both sides.

46x+23−23=−23−23

46x=−46

Step 4: Divide both sides by 46.

46x/46 = -46/46

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3 years ago

Con 5,2l tenemos para 100km, con 60l para cuantos km tenemos?

liraira [26]

Answer:

Step-by-step explanation:

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2 years ago

ABCD is a square of side 3, and E and F are the mid points of sides AB and BC respectively. What is the area of the quadrilatera

Dima020 [189]

Given : - Square ABCD with side 3. E and F as midpoints.
To find : - area of EBFD

Solution : - We have, area of square ABCD = 3 x 3 = 9 units.

Thus, (ar)EBFD = ar ABCD - ar DAE - arDCF

arDAE = 1/2 x base x height

=1/2 x 1.5 x 3 ( AE is 1/2 of AB = 1.5, DA is altitude)
= 2.25

arDFC = 1/2 x base x height
= 1/2 x 1.5 x 3 (FC is 1/2 of BC, DC is altitude)
= 2.25

Thus, (ar) EBFD = arABCD - arDAE - arDCF

= 9 - 2.25 - 2.25

= 4.5 units.

Thus, area of quad EBFD is 4.5 units.

7 0

3 years ago

Verify that the divergence theorem is true for the vector field f on the region

schepotkina [342]

$\mathbf f(x,y,z)=5x\,\mathbf i+xy\,\mathbf j+2xz\,\mathbf k$

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By the divergence theorem, the flux of $\mathbf f$ across the boundary of $E$ is given by

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_E\mathrm{div}(\mathbf f)\,\mathrm dV$

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4 0

3 years ago