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2 years ago

Please help me with this question!!

Mathematics

1 answer:

quester [9]2 years ago

8 0

Answer:

Step-by-step explanation:

a1 = 6

a2 = 10

a3 = 14

The next member of the sequence is 4 more than the current sequence. Therefore d = 4

a1 = 6

d = 4

n = 13

an = a1 + (n - 1)*d

an = 6 + (n - 1)*4

a_13 = 6 + 12*4

a_13 = 6 + 48

a_13 = 54

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I don’t understand this

mylen [45]

To solve for an unknown number, you have to keep isolating it till it is all alone:
A=P(1+rt)
(1+rt)=A/P
rt=A/P -1
r=(A/P-1)/t

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2 years ago

Help fast will give brainiest when 2 people answer but you have to show your work hurry help fast

Vika [28.1K]

Answer:

I don't see any attachment!

Step-by-step explanation:

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2 years ago

How to find variables of these #2 & #3

Lubov Fominskaja [6]

So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles like $\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
\\\\\\
\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
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\cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}$

now.. for 3) will be the second picture below

$\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
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\textit{now, because you already know what "x" is, we can use it below}
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\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
\\\\\\
\textit{and let us use "x" again below}
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\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}$

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3 years ago

Decimals that have the same value are

shepuryov [24]

<span>Decimals that have the same value are equivalent decimals. Equivalent decimals are decimal numbers that have the same value (the same amount). In other words equivalent decimal numbers have the same value but different number of decimals. For example:
0.5 and 0.50 are equivalent decimals.
0.5 and 0.500
0.05 and 0.0500

</span>

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3 years ago

∆ABC is a right angled at B and D is a point on BC if AD = 18cm BD = 9cm and CD =4cm Find AC

Gemiola [76]

In triangle, ABD,

AD²= AB²+BD²

AB² = AD²-BD²

AB² = 18²-9² = 324-81 = 243

AB = √243

In triangle, ABC,

AC² = AB²+BC²

AC² = (√243)²+(13)²

AC² = 243+169

AC = √412

AC = 20.29

3 0

3 years ago