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andreev551 [17]
3 years ago
5

Someone please help me with all !

Mathematics
2 answers:
Arisa [49]3 years ago
8 0

Answer:

109 is not a term

Step-by-step explanation:

goldenfox [79]3 years ago
4 0

You can observe that every next term is computed by adding 4 to the previous one.

This implies that the difference between consecutive terms is always 4, and this is an arithmetic sequence.

In order to find the explicit equation, we can explore the pattern:

a_1 = -1 = -1+4\cdot 0

a_2 = 3 = -1+4\cdot 1

a_3 = 7 = -1+4\cdot 2

a_4 = 11 = -1+4\cdot 3

So, as you can see, we have

a_n = -1+4(n-1) = 4n-5

If 109 is a term of the sequence, there must exist a natural number n such that

109 = 4n-5 \iff 4n = 114 \iff n = \dfrac{114}{4} = 28.5

So, 109 is not a term of the sequence.

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Step-by-step explanation:

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Simplify the Following: <br>⅔ + 2³ - ⅓<br><br>a.) 6 ⅓ <br>b.) 8 ⅓<br>c.) ⁷/2 <br>d.) -6​
allochka39001 [22]

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Step-by-step explanation:

You have to evaluate the power and calculate the sum or difference in order to get the answer

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Simplify by combining like terms: 1.9h + 4.5h + 8
klio [65]

Answer:

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Step-by-step explanation:

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3 years ago
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Evaluate the double integral.
Fynjy0 [20]

Answer:

\iint_D 8y^2 \ dA = \dfrac{88}{3}

Step-by-step explanation:

The equation of the line through the point (x_o,y_o) & (x_1,y_1) can be represented by:

y-y_o = m(x - x_o)

Making m the subject;

m = \dfrac{y_1 - y_0}{x_1-x_0}

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

m= \dfrac{2-1}{1-0}

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

m = \dfrac{1-2}{4-1}

m = \dfrac{-1}{3}

∴

y-2 = -\dfrac{1}{3}(x-1)

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg)   dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ([y^2(-4y+8)] \bigg ) \ dy

\iint_D 8y^2 \ dA =8 \int^2_1  \bigg ( -4y^3+8y^2 \bigg ) \ dy

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\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{-45+56}{3}\bigg]

\iint_D 8y^2 \ dA =8 \bigg [  \dfrac{11}{3}\bigg]

\iint_D 8y^2 \ dA = \dfrac{88}{3}

4 0
3 years ago
In AKLM, m = 1.2 cm, k = 5.1 cm and
Elodia [21]

Answer:

  4.2 cm

Step-by-step explanation:

The law of cosines is applicable.

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  l ≈ √17.4236

  l ≈ 4.2 . . . cm

4 0
3 years ago
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