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2 years ago

Which equation has a graph that lies entirely above the x-axis?

Mathematics

2 answers:

Alinara [238K]2 years ago

8 0

Answer:

(Option C) y = (x – 7)2 + 7

Step-by-step explanation:

hope this helps!! have an amazing day <3

kogti [31]2 years ago

4 0

Answer:

Answer:

Option C.

Step-by-step explanation:

The vertex form of a parabola is

where, a is constant, (h,k) is vertex.

If a<0, then it is a downward parabola and if a>0, then it is an upward parabola.

A downward parabola never lies entirely above the x-axis.

First equation is

It is a downward parabola and vertex is (-7,7).

Second equation is

It is an upward parabola and vertex is (7,-7).

Third equation is

It is an upward parabola and vertex is (7,7).

Fourth equation is

It is a linear equation with y-intercept -7.

Only equation 3 is an upward parabola whose vertex lies above the x-axis.

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What are the possible values of x in 8x2 + 4x = -1?

Westkost [7]

Answer:

The possible values of x are:

x= $\frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18$

and

x= $\frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68$

Step-by-step explanation:

$8x^2+4x+1=0$

Using Quadratic formula to solve this equation:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a = 8 \,\, b = 4\,\, c = -1\\Putting \,\, values \,\, in \,\, the\,\, equation\\x= \frac{-4\pm\sqrt{(4)^2-4(8)(-1)}}{2(8)}\\x= \frac{-4\pm\sqrt{16+32}}{16}\\x= \frac{-4\pm\sqrt{48}}{16}\\x= \frac{-4+ \sqrt{48}}{16} \,\, and \,\, x= \frac{-4- \sqrt{48}}{16}\\x= \frac{-1+ \sqrt{3}}{4} \,\, and \,\, x= \frac{-1- \sqrt{3}}{4}$

The possible values of x are:

x= $\frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18$

and

x= $\frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68$

4 0

3 years ago

A right triangle has a base that measure 39 inches and a height that measures 80 inches. What is the length of its hypotenuse?

guajiro [1.7K]

The Pythagorean Theorum formula is a(squared) + b(squared)= c(squared). If you plug in the numbers, the equation is 39(squared) + 80(squared)= c(squared). That simplifies to 1521+6400=c(squared). Adding results in 7921=c(squared). Finally, when you take the square root of both sides, you are left with 89=c (aka the hypotenuse)

5 0

3 years ago

Which of the following is incorrect?

jeyben [28]

C is wrong...............

8 0

3 years ago

Read 2 more answers

At time t=3, there are 2.5 cubic feet of water in the tub. Write an equation for the locally linear approximation of W at t=3, a

velikii [3]

Answer:

W=0.8333*t

2.917 ft³

Step-by-step explanation:

At time t=3, there is 2.5 ft³ of water in the tub

Finding the relation will be;

3=2.5

1=?

Rate=0.8333 ft³/min

W=0.8333 ft³ / min

W=0.8333*t

At t=3.5 will be;

1 min = 0.8333

3.5 min =?

Apply cross-production

3.5*0.8333 =2.917 ft³

4 0

3 years ago

Read 2 more answers

Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). Pease show how!

mote1985 [20]

Check the picture below.

using a second for it, since it's a horizontal line, then say hmmm we'll use the y-intercept, or (0, 1), so the equation of a line that passes through (2,1) and (0,1)

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
&({{ 0}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-1}{0-2}\implies \cfrac{0}{-2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=0(x-2)$

5 0

3 years ago