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Oduvanchick [21]
3 years ago
6

What is negative 2 times nine

Mathematics
2 answers:
emmainna [20.7K]3 years ago
8 0
-18
try it on mentally or calcular hope its useful
MrMuchimi3 years ago
5 0
-18 because a negative number times positive is  negative.
~JZ
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1. Let f(x) = 5x + 3 and g(x) = 3x<br> What is (f + g)(x)?
mina [271]

Answer:

8x + 3

Step-by-step explanation:

In the picture.

I hope that it's a clear solution.

7 0
3 years ago
Find two whole numbers with a sum of 15 and a product of 54
likoan [24]
A+b=15----(1)
a*b=54----(2)

(1) a=15-b ----(3)

put(3) in(2)

(15-b)*b=54
15b-b^2=54
b^2-15b+54=0
(b-9)(b-6)=0
so b=9 or b=6

if b=9 then a=15-9=6
if b=6 then a=15-6=9

answer two numbers are 6 and 9
3 0
3 years ago
f(x) = 2<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex
loris [4]

Answer:

No answer is possible

Step-by-step explanation:

First, we can identify what the parabola looks like.

A parabola of form ax²+bx+c opens upward if a > 0 and downward if a < 0. The a is what the x² is multiplied by, and in this case, it is positive 2. Therefore, this parabola opens upward.

Next, the vertex of a parabola is equal to -b/(2a). Here, b (what x is multiplied by) is 1 and a =2, so -b/(2a) = -1/4 = -0.25.

This means that the parabola opens upward, and is going down until it reaches the vertex of x=-0.25 and up after that point. Graphing the function confirms this.

Given these, we can then solve for when the endpoints of the interval are reached and go from there.

The first endpoint in -2 ≤ f(x) ≤ 16 is f(x) = 2. Therefore, we can solve for f(x)=-2 by saying

2x²+x-4 = -2

add 2 to both sides to put everything on one side into a quadratic formula

2x²+x-2 = 0

To factor this, we first can identify, in ax²+bx+c, that a=2, b=1, and c=-2. We must find two values that add up to b=1 and multiply to c*a = -2  * 2 = -4. As (2,-2), (4,-1), and (-1,4) are the only integer values that multiply to -4, this will not work. We must apply the quadratic formula, so

x= (-b ± √(b²-4ac))/(2a)

x = (-1 ± √(1-(-4*2*2)))/(2*2)

= (-1 ± √(1+16))/4

= (-1 ± √17) / 4

when f(x) = -2

Next, we can solve for when f(x) = 16

2x²+x-4 = 16

subtract 16 from both sides to make this a quadratic equation

2x²+x-20 = 0

To factor, we must find two values that multiply to -40 and add up to 1. Nothing seems to work here in terms of whole numbers, so we can apply the quadratic formula, so

x = (-1 ± √(1-(-20*2*4)))/(2*2)

= (-1 ± √(1+160))/4

= (-1 ± √161)/4

Our two values of f(x) = -2 are (-1 ± √17) / 4 and our two values of f(x) = 16 are (-1 ± √161)/4 . Our vertex is at x=-0.25, so all values less than that are going down and all values greater than that are going up. We can notice that

(-1 - √17)/4 ≈ -1.3 and (-1-√161)/4 ≈ -3.4 are less than that value, while (-1+√17)/4 ≈ 0.8 and (-1+√161)/4 ≈ 2.9 are greater than that value. This means that when −2 ≤ f(x) ≤ 16 , we have two ranges -- from -3.4 to -1.3 and from 0.8 to 2.9 . Between -1.3 and 0.8, the function goes down then up, with all values less than f(x)=-2. Below -3.4 and above 2.9, all values are greater than f(x) = 16. One thing we can notice is that both ranges have a difference of approximately 2.1 between its high and low x values. The question asks for a value of a where a ≤ x ≤ a+3. As the difference between the high and low values are only 2.1, it would be impossible to have a range of greater than that.

7 0
2 years ago
Solve the system of linear equations. separate the x- and y- values with a coma. 20x=-58-2y
lianna [129]
The best way to solve is by using elimination method.
20x = -58 - 2y
17x = -49 - 2y
Multiply second equation by -1
20x = -58 - 2y
-17x = 49 + 2y
Add equations.
3x = -9 
Divide.
x = -3
Plug in -3 into one of the equations.
17(-3) = -49 - 2y
-51 = -49 - 2y
Add 49 to both sides.
-2 = -2y
Divide.
1 = y
So your solution is (-3, 1).
I hope this helps love! :)

6 0
3 years ago
Write y-6=-2x in slope intercept form
Vera_Pavlovna [14]

Slope:  2

Y-Intercept:   6

2,6

8 0
3 years ago
Read 2 more answers
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