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2 years ago

Work out the value of (6.8x10^2) x (1.3x10^-3)

Mathematics

1 answer:

nikitadnepr [17]2 years ago

6 0

Answer:

0.884

Step-by-step explanation:

6.8x10^2 = 680

1.3x10^-3 = 0.0013

680 X 0.0013 = 0.884

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Tara owes $14,375 in credit card debt. The interest accrues at a rate of 5.3%. She is also borrowing $570 each month for rent fr

siniylev [52]

Answer:

$(f + g)(t) = f(t) + g(t) = 14375 (1 + \frac{5.3}{100})^{t} + 6840t$

$29619.13

Step-by-step explanation:

a. Tara has $14375 in credit card debt and the interest rate is 5.3%.

Now, if f(t) represent the amount of money Tara have in credit card debt, where t is the number of years after after interest begins to accrue, then

$f(t) = 14375 (1 + \frac{5.3}{100})^{t}$ ......... (1)

Again Tara borrows $570 each month for rent from her parents without any interest.

If g(x) represent the amount of money Tara owes to her parents, where t represents the number of years passed,then we can write

g(t) = 570 × 12t = 6840t ........ (2)

Therefore, $(f + g)(t) = f(t) + g(t) = 14375 (1 + \frac{5.3}{100})^{t} + 6840t$

b. So, for t = 2 years,

$(f + g)(t) = 14375 (1 + \frac{5.3}{100})^{2} + 6840 \times 2$ = $29619.13

So, Tara has to repay $29619.13 if she continues this way without any repayment for 2 years. (Answer)

7 0

1 year ago

Darcie wants to crochet a minimum of 3 blankets. Darcie crochets at a rate of 1/15 of a blanket a day. She has 60 days until she

NNADVOKAT [17]

Answer:

The inequality to determine the number of days, s, Darcie can skip crocheting and still meet her goal can be given as:

$s\leq 15$

Step-by-step explanation:

The complete question is:

Darcie wants to crochet a minimum of 3 blankets to donate to a homeless shelter. Darcie crochets at a rate of 1/15 of a blanket per day. She has 60 days until when she wants to donate the blankets, but she also wants to skip crocheting some days so she can volunteer in other ways. Write an inequality to determine the number of days, s, Darcie can skip crocheting and still meet her goal.

Solution:

Given:

Darcie wants to crochet a minimum of 3 blankets to donate.

Rate at which she crochet = $\frac{1}{15}$ of a blanket per day.

Maximum number of days she has = 60.

To find the number of days Dancie can skip out of 60 days and still reach her goal.

Let $s$ represent the number of days she can skip.

Number of days left to crochet = $(60-s)$ days

At rate of $\frac{1}{15}$ of a blanket per day, number of blankets Dancie can corchet in $(60-s)$ days can be given as :

⇒ $\frac{1}{15}(60-s)$

Simplifying using distribution.

⇒ $(\frac{1}{15}.60)-(\frac{1}{15}.s)$

⇒ $4-\frac{s}{15}$

Dancie needs to crochet a minimum of 3 blankets to reach her goal.

Thus, the inequality can be given as:

$4-\frac{s}{15}\geq 3$

Solving the inequality for $s$

Subtracting 4 both sides.

$4-4-\frac{s}{15}\geq 3-4$

$-\frac{s}{15}\geq -1$

Multiplying both sides by -15.

$-15(-\frac{s}{15})\leq -15(-1)$ [On multiplying by a negative number the sign of inequality reverse]

∴ $s\leq 15$

Thus, Dancie can skip a maximum of 15 days.

5 0

2 years ago

at the track meet Jacob and Daniel compete in the 220m- hurdles Daniel finishes in 3 fourths of a minute. Jacob finishes with 5

jeka94

Daniel because 3/4 is larger than 5/12 if you cross multiply.

8 0

2 years ago

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2

Margaret [11]

<u>Answer-</u>

$\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}$

<u>Solution-</u>

Given expression is $(x+2)^4$

Applying Binomial Theorem

$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here,

a = x, b = 2 and n = 4

So,

$\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i$

Expanding the summation

$=\dfrac{4!}{0!\left(4-0\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(4-1\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(4-2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(4-3\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(4-4\right)!}x^0\cdot \:2^4$

$=\dfrac{4!}{0!\left(4\right)!}x^4\cdot \:2^0+\dfrac{4!}{1!\left(3\right)!}x^3\cdot \:2^1+\dfrac{4!}{2!\left(2\right)!}x^2\cdot \:2^2+\dfrac{4!}{3!\left(1\right)!}x^1\cdot \:2^3+\dfrac{4!}{4!\left(0\right)!}x^0\cdot \:2^4$