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3 years ago

Who likes me the best out of the dragon ball universe?

Mathematics

2 answers:

leonid [27]3 years ago

4 0

Answer:

meeeee

Step-by-step explanation:

Evgen [1.6K]3 years ago

4 0

Answer:

i love it meee

Step-by-step explanation:

You might be interested in

Use the functions to answer the question.

Ksivusya [100]

Composition function rule (f○g)(x) = f(g(x))
Given the separate functions: $f(x)=2x+1$ and $g(x)=x^2$

$(f\circ g)(-5)\rightarrow\text{ Composition Function Rule}\rightarrow f(g(-5))=2((5)^2)+1$
$\rightarrow2(25)+1=50+1=51$

4 0

3 years ago

3(12+b)=3x12+3xb

Vikentia [17]

This is the distributive property. You take what is outside of the parenthesis and DISTRIBUTE it inside the parenthesis.

4 0

3 years ago

Ach equation: - 5(– 2x - 4) + 5x - 4 =- 29

Zinaida [17]

- Multiply (-2x-4) by -5:

[(-5)(-2x) + (-5)(-4)] +5x - 4 = -29

= 10x+20+5x-4=-29

- Combine Like Terms:

(10x+5x) + (20-4) = -29

15x+16=-29

- Subtract 16 from each side

15x+16 -16 = -29 -16

15x = -45

x = -3

3 0

3 years ago

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)$

$Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)$

$Midpoint=\left(-4,1\right)$

Slope of lines NY is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{-3-5}{3-(-11)}$

$m=\dfrac{-8}{14}$

$m=\dfrac{-4}{7}$

Product of slopes of two perpendicular lines is -1. So,

$m_1\times \dfrac{-4}{7}=-1$

$m_1=\dfrac{7}{4}$

The perpendicular bisector of NY passes through (-4,1) with slope $\dfrac{7}{4}$ . So, the equation of perpendicular bisector of NY is

$y-y_1=m_1(x-x_1)$

$y-1=\dfrac{7}{4}(x-(-4))$

$y-1=\dfrac{7}{4}(x+4)$

$y-1=\dfrac{7}{4}x+7$

Add 1 on both sides.

$y=\dfrac{7}{4}x+8$

Therefore, the equation of perpendicular bisector of NY is $y=\dfrac{7}{4}x+8$ .

6 0

2 years ago

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

Hence final answer is -0.47.

7 0

3 years ago