Which statements about the graph of y=tax are true

Suppose two bicyclists start at the same location.

DerKrebs [107]

Answer:

Step-by-step explanation:

See attachment.

If both cyclists travel for the same time and speed, they will have travelled the same distance. Since one is headed north and the other east, we can see that the distance between them in one hour is the hypotenuse of a right triangle. Each leg has distance x. We can say x^2 + x^2 = (3 $\sqrt{2}$ )^2

2x^2 =1 8

x^2 = 9

x = 3

They both rode 3 miles.

5 0

2 years ago

Enrique and Yolanda collected 500 pieces of candy on Halloween. They gave half of their candy to charity. Yolanda took 180 piece

777dan777 [17]

The answer would be C. (He counted Yolanda's candy as his own).

This is found by multiplying 500 (starting number of candy) and .64 (percentage divided by a hundred). Thjs would guve you 320, which you would then subtract from the starting number of candy (500) to get 180. 180 is Yolanda's number of candy, which gives you the answer.

4 0

2 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>

7 0

3 years ago

Giá trị x thoả mãn x(x − 3) + x − 3 =0 là

svet-max [94.6K]

Here we will use the trial and error method.

We will try putting different values of x.

$\begin{gathered}\\ \sf\longmapsto 1(1-3)+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto (-2)+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2+1-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2-2=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -4≠0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2(2-3)+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2(-1)+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -2+2-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -3≠0\end{gathered}$

<h2>Again,</h2><h3>x=3</h3>

$\begin{gathered}\\ \sf\longmapsto 3(3-3)+3-3=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 3(3-3)=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 3×0=0\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 0=0\end{gathered}$

<h3>☣Hence, The value of X as 3 satisfies the equation!</h3>

5 0

2 years ago

What is the leading coefficient of the function f(x)= 3x^5+6x^4-x-3

Natali5045456 [20]

3. The leading coefficient of the function f(x)= 3x⁵+6x⁴-x-3 is 3.

For the function f(x)= 3x⁵+6x⁴-x-3 , the highest power of x is 5, so the degree is 5. The leading term is the term containing that degree, 3x⁵. The leading coefficient is the coefficient of that term, 3.

4 0

2 years ago