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katrin2010 [14]
3 years ago
11

PLZ HELP ME!!! Plzzzzzzzzz. ¯\_(ツ)_/¯

Mathematics
1 answer:
PolarNik [594]3 years ago
3 0

Answer:

The digits are 37 (Thirty seven).

Step-by-step explanation:

Given:

Two digit number.

Let the unit place digit be y

and

The tens place digit be (y - 4) as it is 4 less than the unit digit.

So the two digit number is (y-4)(y).

According to question:

Seven times the tens digit plus five times the units digit is equal to 56.

⇒  7(y-4)+5(y)=56

⇒  7y-28+5y=56

⇒   12y=56+28   ...  both sides subtarcting with 28

⇒  12y=84

⇒   y=\frac{84}{12}=7, as\ 12*7=84      ...both sides dividing with 12

So the ones place digit is y=7 and tens place digit is (y-4)=(7-4)=3

The number can be arranged as, from right side we have to put the unit digit and then the tens place digit.

So the digits are 37 in arranged order fulfilling our criteria.

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Select the fraction equivalent of 0.06. reduce to the lowest terms
viva [34]

Answer: 3/50

Step-by-step explanation:

0.06 = 6/100 , 100 would be the denominator because we have two figures after the decimal point. Each figures can also be represented by 10,

Again,

0.06 = 6 × 10-²

Now 0.06 = 6/100

= 3/50.

Therefore, the fractional form = 3/50 in its lowest term.

3 0
3 years ago
There is a box with an area of 112 sq inches and a height of 3.5 inches. What is the width of the box?
Rashid [163]

Answer:

32 inches

Step-by-step explanation:

Because height times width equals area,

we can divide 112 by 3.5 to get the width

So it is 32 inches

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3 years ago
HELP PLEASE!!! HURRY!!! EXPLAIN YOUR ANSWER!!Which of the following rational expressions can be simplified?
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They all technically can be but I think it’s b
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Polygon ABCDEF is translated 12 units to the left to create polygon A′B′C′D′E′F′. If DE = 7 units, what is D′E′ ?
gizmo_the_mogwai [7]
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Read 2 more answers
Y = 1/(x² + c)
Gennadij [26K]

Answer:

y(x)=\frac{1}{x^2-1}

x\in (1,\infty)

Step-by-step explanation:

We are given that one parameter family of solution of the first order DE

y=\frac{1}{x^2+c}

DE

y'+2xy^2=0

y(2)=\frac{1}{3}

Substitute x=2

\frac{1}{3}=\frac{1}{2^2+c}

\frac{1}{3}=\frac{1}{4+c}

4+c=3

c=3-4=-1

Substitute the value of c

y(x)=\frac{1}{x^2-1}

Solution is defined for all  values of x except x=\pm 1

Therefore, by unique existence theorem

The largest interval on which the solution is defined

x\in (1,\infty)

7 0
3 years ago
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