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3 years ago

Jerry deposits $80.00 into a new savings account.

Mathematics

1 answer:

Temka [501]3 years ago

6 0

Answer:

91.80

Step-by-step explanation:

You can get this simple answer by finding the number value of 3.5%, which would be .035. Since you are trying to go up by 3.5%, you can take the .035 and add a 1 onto it to signify that you are not losing any money, you are gaining it or staying the same at least. If you want to see how much money he will have after 4 years, just multiply 80 by 1.035, then take the answer and multiply it by 1.035 again. Then take the answer and do it again to signify the third year, and then take that answer and multiply it by 1.035 to signify the fourth and final year. You will see that he gets about $91.80 after 4 years.

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1/2(-9 + y)=2.5 solve for why. Please show me your steps

Oksi-84 [34.3K]

Answer:

Y = 14

Step-by-step explanation:

1/2(-9 + y) = 2.5

1/2 * -9 = -4.5

1/2 * y = 0.5y

-4.5 + 0.5y = 2.5

+4.5 +4.5

0.5y = 7

/0.5 /0.5

y = 14

7 0

2 years ago

Read 2 more answers

Ruth borrowed a book from the library. The library charged a fixed rental for the book and a late fee for every day the book was

Wewaii [24]

Answer:

The late fee per day for the book

Step-by-step explanation:

2 + .25x

The 2 is the fixed rental for the book

.25x is the late fee per day where .25 is the charge per day and x is the number of days

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3 years ago

Explain why S = {(5, 4, -2), (-15, -12, 6), (10, 8, -4)} is NOT a basis for R^3 (1 point Sis linearly dependent and spans R3 S i

earnstyle [38]

$S$ would be a basis for $\mathbb R^3$ if

(1) the vectors in $S$ are independent, and

(2) the vectors span $\mathbb R^3$ .

Linear independence requires that $c_1=c_2=c_3=0$ is the only solution to

$c_1(5,4,-2)+c_2(-15,-12,6)+c_3(10,8,-4)=(0,0,0)$

These vectors are not linearly independent because if $c_1=3$ , $c_2=1$ , and $c_3=0$ , we have

$3(5,4,-2)+(-15,-12,6)=(15-15,12-12,-6+6)=(0,0,0)$

so $S$ is not a basis for $\mathbb R^3$ .

7 0

3 years ago

One number is 6 more than anouther number. the sum of their squares is 90

Flura [38]

Let x = greater number
      y = smaller number

(1) $x = y + 6
$
(1) $y = x-6$

(2) $x^{2} + y^{2} = 90$
We'll substitute y in (1) to (2)
(2) $x^{2} + (x-6)^{2} = 90$
   $x^{2} + (x^{2} - 12x + 36) = 90$
   $x^2 + x^2 -12x + 36 - 90 = 0$
   $2x^2 - 12x -54 = 0
$
   $2(x^2 - 6x - 27) = 0$
   $2(x - 9)(x + 3) = 0$
x - 9 = 0 or x + 3 = 0
      x = 9 x = -3
and
y = x - 6 y = x - 6
y = 9 - 6 y = -3 - 6
y = 3 y = -9

Therefore, the two numbers can be 9 and 3 or -3 and -9.

6 0

3 years ago

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Marrrta [24]

ASA should be the answer

8 0

3 years ago