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Ilya [14]
3 years ago
15

Leo's model airplane uses O.03 L of fuel each minute it flies.Lf the fuel tank holds O.5 L, how long can the plane fly without r

efueling? (Round to the nearest O.1 minute.)
Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

16 minutes

0.03*16=0.48

Closest you can get w/out going over.

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You have 2\3 of a pizza left from yesterday. You divide it into 4 equal pieces. What fraction of the pizza is each piece?​
romanna [79]
<h3>Answer:  1/6</h3>

====================================================

Explanation:

Let's say for example this pizza had 6 slices to start with. Two-thirds of this is (2/3)*12 = 4 slices. So you have 4 slices left over from yesterday. Then let's say you hand one slice to each of your four friends. Each friend would have 1/6 of the original pizza.

4 0
2 years ago
Question in picture above ^
Andrei [34K]
[ 4(4) x 4(3) ] / 4(5)

= [4(4+3)] / 4(5)

= 4(7) / 4(5)

= 4(7-5)

= 4(2)
5 0
2 years ago
Darius is a graphic designer he makes a sign for a company that has a width of 2 feet and a length of 2 feet 6 inches the compan
Sav [38]

Answer:

\frac{1}{3}

Step-by-step explanation:

Width of the sign made by Darius initially = 2 feet

Length of the sign = 2 feet 6 inches

Dimension of the flyer are 10 inches by 8 inches. Since, in original sign the measure of length is greater, therefore, the length of flyer is 10 inches and its width is 8 inches.

In order to find the scale factor we must convert the lengths and widths to same units. Lets convert the length and width of sign into inches.

Since, 1 feet = 12 inches

Length of the sign = 2 feet 6 inches = 2(12) + 6 inches = 30 inches

Width of the sign = 2 feet = 2(12) inches = 24 inches

Now we can find the scale factor by either comparing the lengths or widths of both the designs. Scale factor will be equal to the ratio of corresponding lengths/widths.

So, the scale factor would be:

Length of Flyer : Length of Sign

= 10 inches : 30 inches

= 1 : 3

= \frac{1}{3}

This shows, the length of flyer is \frac{1}{3} times as that of the sign. So, the scale factor that Darius must use is \frac{1}{3}. The length and width of the flyer are \frac{1}{3} as that of the sign.

The same scale factor would result if we would have used the ratio of widths instead of the lengths.

Scale Factor = Width of Flyer : Width of sign

= 8 : 24

= 1 : 3

Therefore, Darius must type the scale factor of \frac{1}{3} in his computer to get the size of the flyer.

7 0
2 years ago
If JK answered / of the Math questions correctly, how many points did she get in a 50- item test?
Fittoniya [83]

Answer:

JK got 30 answers correct on a 50 item test.

Step-by-step explanation:

It is given that:

Total items on test = T = 50

Fraction of total items that were answered correctly = 3/5

We simply have to multiply the total number of questions by the fraction of correct answers.

So,

The correct answered number of items are:

= \frac{3}{5} * 50\\=30

Hence,

JK got 30 answers correct on a 50 item test.

4 0
3 years ago
(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2
SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(c) y_p is -4x + 7

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

(c) y'' + 2y = −8x + 14

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 16x + 7

2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

8 0
3 years ago
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