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3 years ago

15

If each muffin had the same amount of sugar, how many kilograms of sugar, to the nearest thousand are in each corn muffin ( 1,00

0 Muffins ) ( 63.2 kilograms )

1 answer:

Tresset [83]3 years ago

5 0

Answer:

Sorry bro I didn't understand

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Julia is allowed to watch no more than 5 hours of television a week. So far this week, she has watched 1.5 hours. Write and solv

Leona [35]

The inequality is used to solve how many hours of television Julia can still watch this week is $x + 1.5 \leq 5$

The remaining hours of TV Julia can watch this week can be expressed is 3.5 hours

<h3><u>Solution:</u></h3>

Given that Julia is allowed to watch no more than 5 hours of television a week

So far this week, she has watched 1.5 hours

To find: number of hours Julia can still watch this week

<em>Let "x" be the number of hours Julia can still watch television this week</em>

"no more than 5" means less than or equal to 5 ( ≤ 5 )

Juila has already watched 1.5 hours. So we can add 1.5 hours and number of hours Julia can still watch television this week which is less than or equal to 5 hours

number of hours Julia can still watch television this week + already watched ≤ Total hours Juila can watch

$x + 1.5 \leq5$

Thus the above inequality is used to solve how many hours of television Julia can still watch this week.

Solving the inequality,

$x + 1.5 \leq5\\\\x \leq 5 - 1.5\\\\x \leq 3.5$

Thus Julia still can watch Television for 3.5 hours

5 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.

a)

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:

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

b)

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

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

c)

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.

d)

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

e)

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

3 years ago

In circle B with the measure of minor arc AC= 84°, find m

ELEN [110]

Answer:

m<ADC = 42°

Step-by-step explanation:

Based on the inscribed angle theorem, the inscribed angle, m<ADC is ½ the measure of minor arc AC.

If minor arc is given as 84°, therefore:

m<ADC = ½(84)

m<ADC = 42°

3 0

2 years ago

At 11560 ft above sea level climax Colorado is the highest town in the USA. The lowest town is calipatria, California at 185 ft

Ugo [173]

The first integer is 11560 and the other one is -185 Calipatria is 11375 feet closer to sea level Please mark as brainliest if you like my answer :D

3 0

3 years ago

Read 2 more answers

I need help on 9-11 please

lisabon 2012 [21]

Hello! I would love to help!

9. Basically, it is asking us to multiply 5.2 by 46.

5.2*46=239.2

So, your answer is 239.2 inches.

10. First, let's find out how many blocks she has left to run by subtracting the number of blocks she has already ran by the number of total blocks.

60-18=42

So Susan has 42 blocks left to run. Now, let's divide that by 2 since she can run 2 blocks per minute.

42/2=21

So, the answer is 21 minutes.

11. Let's set up an equation: We know that she will automatically pay $49.95 per month, plus $0.15 per additional minute. We also know her final total for the month was $61.20:

49.95+0.15m=61.20

Let's solve! Subtract both sides by 49.95

49.95+0.15m-49.95=61.20-49.95

Simplify

0.15m=11.25

Now, divide both sides by 0.15

0.15m/0.15=11.25/0.15

M=75

So your answer is 75 additional minutes.

Hope this helped! Comment if you have questions!

5 0

3 years ago