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BaLLatris [955]

2 years ago

15

The table below shows the lunch

1 answer:

ss7ja [257]2 years ago

6 0

0.48$+2*1.89$+1.29$+0.99$=6.54$
10$-6.54$=3.46$
It’s 3.46$ change he’ll get back

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Divide this equation and show your work.(QUICKLY)<br> X^3+4x^2-25x-28/x-4

xxTIMURxx [149]

hope this helps

andddd yeaaaaaaa hope !!!!!!!

8 0

2 years ago

What is the next number 486, 162, 54, 18

Lesechka [4]

The pattern is to divide by 3

486 ÷ 3 = 162
162 ÷ 3 = 54
54 ÷ 3 = 18
18 ÷ 3 = 6

The next number is 6

5 0

2 years ago

Read 2 more answers

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

2 years ago

A fund earns a nominal rate of interest of 6\% compounded every two years. Calculate the amount that must be contributed now to

Nesterboy [21]

Answer:

$712.

Step-by-step explanation:

We have been given that a fund earns a nominal rate of interest of 6% compounded every two years. We are asked to find the amount that must be contributed now to have 1000 at the end of six years.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

$6\%=\frac{6}{100}=0.06$

Since interest is compounded each two years, so number of compounding per year would be 1/2 or 0.5.

$1000=P(1+\frac{0.06}{0.5})^{0.5*6}$

$1000=P(1+0.12)^{3}$

$1000=P(1.12)^{3}$

$1000=P*1.404928$

$\frac{1000}{1.404928}=\frac{P*1.404928}{1.404928}$

$P=711.7802478$

$P\approx 712$

Therefore, an amount of $712 must be contributed now to have 1000 at the end of six years.

7 0

2 years ago

What is the answer to this question? <br> 1.4a^2+4a-63

LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

Δ = $4^{2} - 4 * 4 * (-63) = 16 + 16 * 63 = 16 * (1 + 63) = 16 * 64$

VΔ = 4 * 8 = 32

a = (- 4 ± 32) * $\frac{1}{8}$

a' = 7/2

a''= 9/2

3 0

2 years ago