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3 years ago

I need help ASAP!!!

Mathematics

1 answer:

andre [41]3 years ago

7 0

Answer:

Yes, you do have enough money to make a mosiac. You have $180.26 left.

Step-by-step explanation:

Alright, so we know she has $200 to spend.

The wall to cover is 14 ft x 7 ft (98 sq. ft total)

The window is 6 ft x 5 ft (30 sq. ft total)

Each mosiac tile is 6 in x 6 in (36 sq. in which is 3 sq. ft)

The tiles only come in boxes of 12, so each box covers 36 sq. ft.

(3 x 12 = 36)

The wall we have to cover is the area of the wall - window which is

98 - 30 = 68 sq. ft.

You would need at least 2 boxes to be able to cover the wall

(2 x 36 = 72 sq. ft), which would cost you $19.74.

So yes, you do have enough money to make a mosiac.

You have $180.26 left. ($200 - $19.74)

Hope this helps!

You might be interested in

Prime factorization of 1408

V125BC [204]

Answer:

There are 7 copies of 2 and 1 copy of 11 in the product:

1408 = 2^7×11

Step-by-step explanation:

Factor the following integer:

1408

The last digit of 1408 is 8, which means it is even. Therefore 1408 is divisible by 2:

1408 = 2 704:

1408 = 2×704

The last digit of 704 is 4, which means it is even. Therefore 704 is divisible by 2:

704 = 2 352:

1408 = 2×2×352

The last digit of 352 is 2, which means it is even. Therefore 352 is divisible by 2:

352 = 2 176:

1408 = 2×2×2×176

The last digit of 176 is 6, which means it is even. Therefore 176 is divisible by 2:

176 = 2 88:

1408 = 2×2×2×2×88

The last digit of 88 is 8, which means it is even. Therefore 88 is divisible by 2:

88 = 2 44:

1408 = 2×2×2×2×2×44

The last digit of 44 is 4, which means it is even. Therefore 44 is divisible by 2:

44 = 2 22:

1408 = 2×2×2×2×2×2×22

The last digit of 22 is 2, which means it is even. Therefore 22 is divisible by 2:

22 = 2 11:

1408 = 2×2×2×2×2×2×2×11

11 is not divisible by 2 since 11 is odd and 2 is even:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2)

The sum of the digits of 11 is 1 + 1 = 2, which is not divisible by 3. This means 11 is not divisible by 3:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2 or 3)

The last digit of 11 is not 5 or 0, which means 11 is not divisible by 5:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2, 3 or 5)

Divide 7 into 11:

| | 1 | (quotient)

7 | 1 | 1 |

- | | 7 |

| | 4 | (remainder)

11 is not divisible by 7:

1408 = 2×2×2×2×2×2×2×11 (11 is not divisible by 2, 3, 5 or 7)

No primes less than 11 divide into it. Therefore 11 is prime:

1408 = 2×2×2×2×2×2×2×11

There are 7 copies of 2 and 1 copy of 11 in the product:

Answer: 1408 = 2^7×11

8 0

4 years ago

The Hundred Years War led to the end of the Middle Ages in the beginning of the

Gre4nikov [31]

It led to the beginning of the Renaissance.

4 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

4 years ago

Can someone help me please

Mademuasel [1]

So first of all, both angles are alternate interior angles right. so they would equal the same thing. all you have to do is divide 75 by 3 to get the x which would equal 25. Hope that helps :)

8 0

3 years ago

IV. Intercepts. Give the intercepts of the following linear equations. 1) 5x + 3y = 8 2) 10x + 6y - 12 = 4 3) 2x – 3y = 5 4) 3x

rosijanka [135]

Answer:

1) $\frac{8}{3}$