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oksian1 [2.3K]
3 years ago
5

Tamia’s mother is 3 times older than Tamia. The sum of their age is 52. How old is Tamia and her mom?

Mathematics
1 answer:
Luden [163]3 years ago
4 0
Let’s say that Tamia is x years old. Tamia’s mother is 3x years old. The sum of their ages is 52 years old.
That means

x + 3x = 52
4x = 52
x = 13
Tamia is 13 years old and her mom is 3 times older; 39 years old.
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There are 158 students registered for American History classes. There are twice as many students registered in second period as
Margarita [4]

There are 28 students in first period and 56 students in sceond period and 74 students in third period

<em><u>Solution:</u></em>

Let the number of students in first period be "x"

Let the number of students in second period be "y"

Let the number of students in third period be "z"

<em><u>There are 158 students registered for American History classes.</u></em>

Therefore,

x + y + z = 158 ---------- eqn 1

<em><u>There are twice as many students registered in second period as first period</u></em>

number of students in second period = twice of number of students in first period

y = 2x ------- eqn 2

<em><u>There are 10 less than three times as many students in third period as in first period</u></em>

number of students in third period = 3 times number of students in first period - 10

z = 3x - 10 ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

x + 2x + 3x - 10 = 158

6x = 168

<h3>x = 28</h3>

<em><u>Substitute x = 28 in eqn 2</u></em>

y = 2(28)

<h3>y = 56</h3>

<em><u>Substitute x = 28 in eqn 3</u></em>

z = 3(28) - 10

z = 84 - 10

<h3>z = 74</h3>

Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period

5 0
3 years ago
Find the center of a circle with the equation: x2 y2−32x−60y 1122=0 x 2 y 2 − 32 x − 60 y 1122 = 0
mixas84 [53]

The equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

The center of the circle exists at (16, 30).

<h3>What is the equation of a circle?</h3>

Let, the equation of a circle exists:

$(x-h)^2 + (y-k)^2 = r^2, where (h, k) be the center.

We rewrite the equation and set them equal :

$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0

We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

To learn more about center of the circle refer to:

brainly.com/question/10633821

#SPJ4

7 0
2 years ago
Which set is closed under multiplication?
Sav [38]
I think the answer is D
5 0
3 years ago
Why is (x-3) NOT a factor of x3+2x2-5x-6
ivolga24 [154]

As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Step-by-step explanation:

Given expression is:

x^3+2x^2-5x-6

We have to check if x-3 is a factor of given expression

In order for x-3 to be a factor of expression, we have to put x-3 = 0 => x=3 in expression.

If the expression is zero at x=3 then x-3 is a factor of expression otherwise not.

So putting x=333 in expression

=(3)^3+2(3)^2-5(3)-6\\= 27+2(9) - 15-6\\= 27+18-15-6\\= 45-21\\=24 \neq 0

As the expression is not equal to zero on x=3, x-3 is not a factor of given expression

Keywords: Polynomials, expressions

Learn more about polynomials at:

  • brainly.com/question/4228574
  • brainly.com/question/4279146

#LearnwithBrainly

8 0
4 years ago
Where common multiples between three and five and one and 50
Arlecino [84]
300 is a multiple of 3,5,1, and 50
5 0
3 years ago
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