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maks197457 [2]
3 years ago
8

1) 17, 29, 32, 9, 30, 14, 8, 39, 11, 32, 23 Minimum : Maximum: Q1: Q2: Q3:

Mathematics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

Minimum: 8

Maximum: 39

Q1: 11

Q2: 23

Q3: 32

Step-by-step explanation:

Set the numbers into numerical order

8 9 11 14 17 23 29 30 32 32 39

Then separate into quarters

[8 9 11] [14 17 23] [29 30 32] 32 39

You might be interested in
GEOMETRY!!!!!!!!!!!!!
Rina8888 [55]

(1,F)(2,C)(3,H)(4,A)(5,G)(6,G)(7,D)(8,B)(9,E)

You might want to review the words.

4 0
3 years ago
If x+a is a factor of 2x²+2ax+5x+10,find a.
vazorg [7]

If x+a is a factor of the polynomial, it means that -a is a root of the polynomial.

In fact, if a polynomial p(x) has a root x_0 (i.e. p(x_0)=0), then the polynomial is divisible by (x-x_0)

In your case, you know that the polynomial is divisible by (x+a)=(x-(-a)), so -a is a solution.

Let's evaluate p(-a) and set it to zero:

2(-a)^2+2a(-a)+5(-a)+10 = 0 \iff 2a^2-2a^2-5a+10=0 \iff -5a+10 = 0 \iff a = 2

We can check the answer: if a=2 the polynomial becomes

2x^2+2(2)x+5x+10 = 2x^2+9x+10

And we're claiming that -a=-2 is a root of this polynomial:

2(-2)^2+9(-2)+10 = 2\cdot 4 - 18 + 10 = 0

5 0
3 years ago
Coach Bruch had 12 quarts of milk. How much is this in cups?
Thepotemich [5.8K]

Answer: 48

Step-by-step explanation: there is 4 cups in a quart

5 0
3 years ago
Find X and QT. The shape is a Rhombus
kompoz [17]

Answer:

See below ~

Step-by-step explanation:

Sides of a rhombus are equal.

⇒ QT = TS

⇒ x² - 4x - 10 = 6x + 14

⇒ x² - 10x - 24 = 0

⇒ x² + 2x - 12x - 24 = 0

⇒ x (x + 2) - 12 (x + 2) = 0

⇒ (x + 2)(x - 12) = 0

⇒ x = -2 or x = 12

Substitute both values and see which gives a positive value for QT :

<u>When x = -2</u> :

  • QT = (-2)² - 4(-2) - 10
  • QT = 4 + 8 - 10
  • QT = 2

<u>When x = 12</u> :

  • QT = (12)² - 4(12) - 10
  • QT = 144 - 48 - 10
  • QT = 86

The 2 possible answers are :

  • x = -2, QT = 2
  • x = 12, QT = 86
8 0
2 years ago
If alpha beta are the roots of the equation
Vikki [24]

Answer:

Step-by-step explanation:

Hello, I believe that we can consider a different from 0.

By definition of the roots we can write.

ax^2+bx+c=a(x-\alpha)(x-\beta)=a(x^2-(\alpha + \beta)x+\alpha \cdot \beta)\\\\\text{So we can say that:}\\\\\alpha + \beta = \dfrac{-b}{a}\\\\\alpha \cdot \beta=\dfrac{c}{a}\\\\\text{So the expected quadratic equation is}\\\\(x+\dfrac{b}{a})(x-\dfrac{c}{a})=0\\\\ \Large \boxed{\sf \bf \ (ax+b)(ax-c)=0 \ }

Thank you

8 0
3 years ago
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