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svlad2 [7]
3 years ago
6

7 1/5 + ( 3 2/5 + 5 4/5 ) =​

Mathematics
2 answers:
Levart [38]3 years ago
7 0

Answer:

16 2/5

Step-by-step explanation:

Sophie [7]3 years ago
4 0

Answer:

16\frac{2}{5}

Step-by-step explanation:

7\frac{1}{5}  + 3 \frac{2}{5} +5\frac{4}{5}

1) Add all the whole numbers

2) Add all numerators

3) Check if it's an improper fraction

15\frac{7}{5}  = 16\frac{2}{5}

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Which is greater 2.16 or 2.061
jenyasd209 [6]
2.16 is greater. To do this start from the left and compare the numbers. For example, 2 is the same in both but 1 is greater than 0, so the answer is 2.16.
6 0
3 years ago
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The price of a gym membership has a one-time sign-up fee and a monthly fee. The price can be modeled by the function y = 10x + 1
monitta
15; it represents the one-time sign-up fee

The y-intercept can be found either on the graph where the line intercepts the y-axis, or b in the equation y=mx+b.

It represents the fee because it will be charged even if the number of months (x) is zero. 10 is the monthly fee because it is multiplied by x, the number of months.
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3 years ago
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(3x + 5y = 7<br> { 4x - y = 5
dalvyx [7]

Answer:

Step-by-step explanation:

Solution by substitution method

3x+5y=7

and 4x-y=5

Suppose,

3x+5y=7→(1)

and 4x-y=5→(2)

Taking equation (2), we have

4x-y=5

⇒y=4x-5→(3)

Putting y=4x-5 in equation (1), we get

3x+5y=7

⇒3x+5(4x-5)=7

⇒3x+20x-25=7

⇒23x-25=7

⇒23x=7+25

⇒23x=32

⇒x=32/23

→(4)

Now, Putting x=32/23

in equation (3), we get

y=4x-5

⇒y=4(32/23)-5

⇒y=(128-115)/23

⇒y=13/23

∴y=13/23   and x=32/23

5 0
2 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
Order these three values from least to greatest. Explain or show your reasoning.
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