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3 years ago

Solve this equation with substitution method :- x+5 = 3 (y+5) x-5 = 7 (y-5)

1 answer:

5 0

From the first equation,
x+5 = 3(y+5)
x = 3y + 15 - 5
Now substitue x in the second equation with (3y +15 - 5).
x-5 = 7(y-5)
(3y+15-5) - 5 = 7(y-5)
3y +5 = 7y - 35
-4y = - 40
y = 10

Since y is 10, and x is (3y +15 - 5),
x = 30 + 15 - 5 = 40

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Leap years are years in which February has 29 days instead of 28. The device of leap year was invented to keep the calendar in s

goldenfox [79]

Answer:

As you know, a year has around 365 + 1/4 days.

This means that in two years, we have:

365 + 356 + 1/4 + 1/4 = 730 + 1/2

and so on.

adding this up, when we have 4 years we have a full day extra, this is:

1460 + 1

When we divide 1461 by 4, we have 365 with a surpass of 1.

The rule used to keep the calendar in sync with this extra day is adding an extra day to each fourth year.

So each fourth year, we have an extra day in Februray (the Februray 29th), this is called a bisiest year.

The "math rule" used to know if a year is leap or not is:

if a year is not divisible by 4, then it is a common year

else if the year is not divisible by 100 then it is a leap year,

else if the year is not divisible by 400, then it is a common year

if not, the year is a leap year.

Where "year" represents the number of the year.

8 0

3 years ago

Saturn is 8.867 × 10^8 miles away from the Sun. Uranus is 1.787 × 10^9 miles away from the Sun. Approximately how many times far

Over [174]

Answer:

2 times

Step-by-step explanation:

Please refer to the image attached of our solar system in which Saturn and Uranus are clearly visible.

Distance of Saturn from Sun, $D_{1} = 8.867 \times 10^{8}\ miles$

Distance of Uranus from Sun, $D_{2} = 1.787 \times 10^{9}\ miles$

We have to find, how many times the Uranus is far from Sun than Saturn is far from Sun.

Let us begin by multiplying the distance of Saturn from Sun by 2: $D_{1} \times 2 = 8.867 \times 10^{8}\ miles \times 2\\\Rightarrow 17.734 \times 10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{1} \times 10^{8}\ miles\\\Rightarrow 1.7734 \times 10^{8+1}\ miles\\\Rightarrow 1.7734 \times 10^{9} $\approx$ D_{2}$

From above, we can say that $D_{2}$ is approximately twice of $D_{1}$ .

In other words, we can say that Uranus is at approximately twice distance from Sun than that of Saturn is.

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3 years ago

A mirror frame in the shape of an oval is shown below. The ends of the frame form semicircles: (5 points)

guapka [62]

Answer:

696.265 inches

Step-by-step explanation:

Radius = 27/2 = 13.5

2 semicircles + 2 lengths

(3.14 × 13.5²) + 2(62)

696.265 inches

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3 years ago

Which of the following side lengths are the legs? 6, 14, and 12

Dmitry_Shevchenko [17]

Answer:

The side lengths that are the legs are 6 and 12 because the hypotenuse is the largest side length of a triangle (since there are three side lengths shown).

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aliina [53]

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