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2 years ago

Tas

Mathematics

1 answer:

BigorU [14]2 years ago

3 0

Answer: (8,10)

Step-by-step explanation:

To go from A to C, we go right 2 units and up 6 units.

So, if we let the fourth vertex be D, then to go from B to D, we must also go right 2 units and up 6 units from B.

Therefore, the answer is (8, 10).

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Can someone pls help me with these I’ve asked this question 3 times already :( will mark brainiest

Tresset [83]

1: cell

2: distinction

3: assumption

4: foliage

5: commision

6: viewpoint

7: considerable

8: membrane

9: cell

10: final

11: viewpoint

12: considerable

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Edit: added 9-12 didn't see them until I accidentally scrolled down, lol.

7 0

3 years ago

Read 2 more answers

The right expression to calculate how much money will be in an investment account 14 years from now if you deposit $5,000 now an

Svet_ta [14]

Answer:

The expression to compute the amount in the investment account after 14 years is: <em>FV</em> = [5000 ×(1.10)¹⁴] + [3000 ×(1.10)⁸].

Step-by-step explanation:

The formula to compute the future value is:

$FV=PV[1+\frac{r}{100}]^{n}$

PV = Present value

r = interest rate

n = number of periods.

It is provided that $5,000 were deposited now and $3,000 deposited after 6 years at 10% compound interest. The amount of time the money is invested for is 14 years.

The expression to compute the amount in the investment account after 14 years is,

$FV=5000[1+\frac{10}{100}]^{14}+3000[1+\frac{10}{100}]^{14-6}\\FV=5000[1+0.10]^{14}+3000[1+0.10]^{8}$

The future value is:

$FV=5000[1+0.10]^{14}+3000[1+0.10]^{8}\\=18987.50+6430.77\\=25418.27$

Thus, the expression to compute the amount in the investment account after 14 years is: <em>FV</em> = [5000 ×(1.10)¹⁴] + [3000 ×(1.10)⁸].

4 0

3 years ago

the total length of 4 blue banners an 5 yellow banners is 49 meters. the total length of 2 blue banners an 1 yellow banner is 17

marin [14]

Yellow banner is 5
Blue banner is 6

8 0

3 years ago

Read 2 more answers

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago

HELP ME WITH MATH PLS

Natasha2012 [34]

Answer: y = (xm)/3 + b

Step-by-step explanation:

1. Multiply m on both sides

x × m = 3(y - b)

2. Divide by 3 on both sides

(xm)/3 = y - b

3. Add b on both sides

y = (xm)/3 + b

8 0

3 years ago