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allochka39001 [22]
2 years ago
5

Your friend says that the solutions to the inequality 5x - 14x < - 18 are x ≤2. Solve the inequality correctly. What error di

d your friend
make?
Mathematics
1 answer:
katrin2010 [14]2 years ago
6 0
<h3>Answer:  x > 2</h3>

Work Shown:

5x - 14x < - 18

-9x < - 18

x > -18/(-9)

x > 2

The inequality sign flips when dividing both sides by a negative number.

Here's another approach you could take.

5x - 14x < - 18

-9x < -18

0 < -18+9x

18 < 9x

9x > 18

x > 18/9

x > 2

It's a slightly longer pathway, but it avoids a sign flip when you divide both sides by the positive number 9.

The sign flip happens two steps earlier when going from 18 < 9x to 9x > 18. In other words, A < B is the same as B > A.

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Drag the tiles to the correct boxes to complete the pairs. not all tiles will be used. match each quadratic equation with its so
Oxana [17]

A quadratic equation is an equation whose leading coefficient is of the second degree. The given quadratic equation can be matched with its solution as shown below.

<h3>What is a quadratic equation?</h3>

A quadratic equation is an equation whose leading coefficient is of second degree also the equation has only one unknown while it has 3 unknown numbers. It is written in the form of ax²+bx+c.

The given quadratic equation can be matched with its solution as shown below,

2x² − 32 = 0 → x = √(32/2) → x = {-4, 4}

4x² − 100 = 0 → x = √(100/4) → x = {-5,5}

x² − 55 = 9 → x = √(9+55) → x = {-8,8}

x² − 140 = -19 → x = √(-19+140) → x = {-11,11}

2x² − 18 = 0 → x = √(18/2) → x = {-3, 3}

Learn more about Quadratic Equations:

brainly.com/question/2263981

#SPJ1

7 0
2 years ago
Let z be a bernoulli random variable with parameter p and w an independent binomial random variable with parameters n and p. Wha
Maurinko [17]

Answer:

It is known that if Z is a binomial random variable with parameters n  and p and in addition, W is a binomial random variable, independent of X, with parameters m and p, it is assumed that the variable R = Z + W, is a binomial random variable with parameters (n + m and p. In this case, m = 1. Therefore, R is a binomial random variable with parameters (n + 1) and p.

Step-by-step explanation:

4 0
3 years ago
3.(08.07 HC)
SashulF [63]

Answer:

For a general equation:

y = f(x) =  a*x^2 + b*x + c

The x-intercepts are given by:

x = \frac{-b \pm \sqrt{b^2 - 4*a*c} }{2*a}

And the vertex is the point (h, k) such that:

h = -b/2*a

k = f(h)

Part A)

We have the function f(x) = -16*x^2 + 22*x + 3

The x-intercepts are then:

x = \frac{-(22) \pm \sqrt{22^2 - 4*(-16)*3} }{2*(-16)} = \frac{-22 \pm 26 }{-32}

The two intercepts are:

x = (-22 - 26)/(-32) = 1.5

x = (-22 + 26)/(-32) = -0.125

Part B:

The vertex is (h, k), such that:

h = -22/(2*(-16)) = -22/-32 = 0.6875

k = f(0.6875) = -16*(0.6875)^2 + 22*0.6875 + 3 = 10.6525

Then the vertex is: ( 0.6875, 10.6525)

If the leading coefficient is positive, then the vertex is a minimum

If the leading coefficient is negative, then the vertex is a maximum.

In this case the leading coefficient is -16, then we can conclude that the vertex is a maximum.

Part C:

To graph the function, we can graph the points that we already know (the vertex and the two x-intercepts) and connect them with a curve.

You could also add another few points so you have a guide to draw the curve, for example the point:

y = f(0) = -16*0^2 + 22*0 + 3

    f(0) = 3

(0, 3)

6 0
3 years ago
Answer Fast.............
Aleks04 [339]
1and 1/10 is correct

brainliest or nah lol
4 0
3 years ago
The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func
katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle (\frac{1}{3},\frac{1}4).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A(\frac{1}{3},\frac{1}4)

Let O be the origin i.e. (0,0)

Point B will be (\frac{1}{3},0)

Now, let us consider the right angled triangle \triangle OBA:

Sides:

Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}

Using Pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}

sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}

\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}

cos\angle AOB = \dfrac{Base}{Hypotenuse}

\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}

tan\angle AOB = \dfrac{Perpendicular}{Base}

\Rightarrow tan\angle AOB = \dfrac{3}{4}

cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}

\Rightarrow cosec\angle AOB = \dfrac{5}{3}

sec\angle AOB = \dfrac{Hypotenuse}{Base}

\Rightarrow sec\angle AOB = \dfrac{5}{4}

cot\angle AOB = \dfrac{Base}{Perpendicular}

\Rightarrow cot\angle AOB = \dfrac{4}{3}

3 0
3 years ago
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