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sergij07 [2.7K]

3 years ago

6

Graph the function f(x) = (x + 1)(x – 5). Use the drop-down menus to complete the steps needed to graph the function. Identify t

he x-intercepts: (–1, 0) and (5, 0) Find the midpoint between the intercepts: (2, 0) Find the vertex: Find the y-intercept: Plot another point, then draw the graph.

2 answers:

Sveta_85 [38]3 years ago

7 0

The Vertex is (2,-9)
and the y-intercept is (0,-5)

den301095 [7]3 years ago

3 0

Vertex = 2, -9
Intercept = 0, -5

You can find this by distributing the parenthesis and then finding the vertex using -b/2a.

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Naily [24]

Answer:

you can try adding the adding angles up to 180

or if its graphed you can count the units

Step-by-step explanation:

and know that all four sides are congruent and diagonals are perpendicular

i tried :)

6 0

3 years ago

Can I get some help plz thank u sm

weeeeeb [17]

Answer:

x = 72

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3 years ago

Can someone help me with this please

DIA [1.3K]

Answer:

See below

Step-by-step explanation:

The starting point of 0 has a value of 3. The common difference of 2 is added at each step increse in n.

n f(n)

0 3

1 5

2 7

3 9

4 11

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5 0

3 years ago

Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

3 years ago

Find the value of y in the right triangle below.<br><br> A. 40<br> B. 90<br> C. 30<br> D. 50

zhannawk [14.2K]

Answer:

A. 50

Step-by-step explanation:

Sum of interior angles of a triangle is always =180 °

Since this is a right triangle, therefore we know that one angle of the triangle measures 90°

Therefor sum of other two angles has to be = $180\°-90\°=90\°$

One angle measures = 40°

Therefore the missing angle $y=90\°-40\°=50\°$

$y=50\°$

5 0

3 years ago