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3 years ago

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If the conversion rate for Swiss francs is 1.68 francs = 1 U.S. dollar, then how many Swiss francs can be exchanged for $79.00 i

n U.S. currency?

1 answer:

Likurg_2 [28]3 years ago

3 0

If one U.S. dollar is as much as 1.68 Swiss francs, then 79,00 U.S. dollars will be as much as 79,00 * 1,68, which is 132,72 Swiss francs.

I hope it will help :)

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70 - q - q - 2q = 80???

Aneli [31]

70-q-q-2q= 80

combine like terms first

so you get 70-4q= 80

subtract 70 on both sides and you get -4q= 10

divide -4 on both sides and you get q= 10/-4 or -2.5

3 0

3 years ago

PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

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harkovskaia [24]

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notka56 [123]

Two negatives make a positive so, -13 - (-49) equals -13 + 49, or 49-13, they are all the same thing, all equal to 36.

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dem82 [27]

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