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3 years ago

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

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GEOMETRY!!!!!!!!!!!!!

Rina8888 [55]

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

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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

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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