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2 years ago

Maggie’s brother is 3 years younger than four times her age. The sum of their age is 22 how old is maggie

Mathematics

1 answer:

gayaneshka [121]2 years ago

8 0

Hey there!

Let's write this as a system of equations.

m+b=22

b= 4m-3

Let's plug our value for b into the first equation and solve for m.

m+4m-3=22

We combine like terms.

5m-3=22

We add 3 to both sides.

5m=25

We divide both sides by 5.

m=5

Therefore, Maggie is 5 years old.

I hope this helps!

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find the perimeter of an isosceles triangle with side lengths of 5+x, 5+x, and xy. Write in simplest form.

Zinaida [17]

<h3>Answer: perimeter = 10+2x+xy</h3>

To get the perimeter, you add up all the outer sides

side1+side2+side3 = (5+x)+(5+x)+(xy) = 10+2x+xy

The like terms 5 and 5 add to 10. The other pair of like terms x and x add to 2x. There isn't anything to pair with xy, so we leave it as is.

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2 years ago

The international cricket team is going to stadium to play the T20 cricket game. A coordinate grid is superimposed on a highway

marysya [2.9K]

Answer:

The coordinate of the rest stop is: $(5,2.5)$

The distance between the hotel and the stadium is 32 miles

Step-by-step explanation:

Given

$(x_1,y_1) = (3,4)$ --- Team hotel

$(x_2,y_2) = (7,1)$ --- Stadium

Solving (a): The coordinates of the rest stop

The rest stop is at half way;

So, the coordinate is:

$(x,y) = \frac{1}{2}(x_1+x_2,y_1+y_2)$

This gives:

$(x,y) = \frac{1}{2}(3+7,4+1)$

$(x,y) = \frac{1}{2}(10,5)$

Open bracket

$(x,y) = (5,2.5)$

Solving (b): Distance between the hotel and the stadium

We have:

$(x_1,y_1) = (3,4)$ --- Team hotel

$(x_2,y_2) = (7,1)$ --- Stadium

The distance (d) is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

So, we have:

$d = \sqrt{(3 - 7)^2 + (4 - 1)^2}$

$d = \sqrt{(-4)^2 + 3^2}$

$d = \sqrt{16 + 9}$

$d = \sqrt{25}$

$d =5$

From the question, we have:

$1\ unit = 6.4\ miles$

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$d =5 * 6.4\ miles$

$d =32\ miles$

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2 years ago

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ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

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$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

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$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

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Answer:

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