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gulaghasi [49]
2 years ago
8

Suppose $580 is placed in a savings account at a simple interest rate of

Mathematics
1 answer:
Anna35 [415]2 years ago
3 0

Answer: $95.7

Step-by-step explanation:

580 x 5.5% = 31.9 x 3 =95.7

$95.7

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Ket [755]

Answer:

D

Step-by-step explanation:

If x is not equal to zero, then 1/x can be equal to x.

So which is k * 1/x if x =1?

D, because x is the same as 1x, which is equivalent to 1/1 (x).

I may be wrong though please look at other answers and if you have questions then comment

8 0
2 years ago
Help on this please??
Anuta_ua [19.1K]

Answer:

23

Step-by-step explanation:

Its half of TU(46).

Hope it helps!

3 0
3 years ago
Write a linear equation to describe the number pattern that starts with –8 and continues with negative multiples of 8. PLEEEEEEE
Svetach [21]
D because it the answer for your question because it
4 0
3 years ago
(Round to the nearest tenth of a percent.) In 2011, the IRS increased the deductible mileage cost to
RUDIKE [14]

Answer:

\%Change = 8.8\%

Step-by-step explanation:

Given

Initial = 51\ cents

Final = 55.5\ cents

Required

Determine the percentage change

Percentage change is calculated as;

\%Change = \frac{Final - Initial}{Initial} * 100\%

\%Change = \frac{55.5 - 51 }{51 } * 100\%

\%Change = \frac{4.5}{51} * 100\%

\%Change = \frac{4.5* 100\%}{51}

\%Change = \frac{450\%}{51}

\%Change = 8.8\%

<em>Hence, the percentage change is approximately 8.8%</em>

6 0
2 years ago
Can you answer these please many thanks
givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

(o) s(3s-4)=3s^2-4s        (using a(b-c)=ab-ac)

(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

(q) 4y(x-3)=4xy-12y      (using a(b-c)=ab-ac)

(r) 5a(2b-5)=10ab-25a     (using a(b-c)=ab-ac)

(s) 4a(3b+2c)=12ab+8ac    (using a(b+c)=ab+ac)

(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

6 0
3 years ago
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