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3 years ago

The equation x=6y^2 describes a parabola. Which way does the parabola open

Mathematics

2 answers:

ELEN [110]3 years ago

5 0

If the coefient in front of the squared term is positive, it opens to the right or up
if it is in front of x^2, it opens up
if it is in front of y^2, it opens right

opens to the right

Verizon [17]3 years ago

5 0

Answer:

The parabola opens right.

Step-by-step explanation:

The equation of the parabola is $x=6y^2$

The equation of the horizontal parabola is $x=ay^2$

If a >0 then the parabola open right
And if a<0 then the parabola opens left.

For the given parabola, the value of a is 6, which is greater than zero.

Hence, the parabola opens right.

We can see it from the attached graph also. The vertex of the parabola is at origin and opens right.

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Plzzzzz bruhhhh I need help

Montano1993 [528]

The answer is roughly 103 square feet.

You can get this by first transforming the measurements using the scale factor. To do this, we need to multiply each measurement by 30.

5.5 in * 30 = 165 in

3 in * 30 = 90 in

Now we have the measurements of the room, we can multiply to find the area.

90 in * 165 in = 14850 square inches.

Now to turn this into feet we need to divide by the number of square inches in a square foot (144).

14850 square inches/ 144 = 103 square feet

7 0

3 years ago

The lengths of the sides of a triangle are 17, 8, n. which of the following must be true?

nlexa [21]

It seems that you have missed the necessary options to answer this question, but anyway, hope this is the answer that you are looking for. Given that the lengths of the sides of a triangle are 17, 8 and n, the one that is considered true would be this: 9 ≤ n ≤ 25. Have a great day!

4 0

4 years ago

Create a matrix that is equal to F+G. The first matrix below is named F and the second matrix below is named G. Name the new mat

likoan [24]

F+G:

$F+G=\begin{bmatrix}{-1.8} & {-8.6} & {} \\ {2.85} & {-1.4} & {} \\ {-1.8} & {5.1} & {}\end{bmatrix}+\begin{bmatrix}{1.32} & {-1.9} & {} \\ {2.25} & {0.0} & {} \\ {-6.2} & {1.4} & {}\end{bmatrix}$

Then, add the elements that occupy the same position:

$H=\begin{bmatrix}{-1.8+1.32} & {-8.6+(-1.9)} & {} \\ {2.85+2.25} & {-1.4+0.0} & {} \\ {-1.8+(-6.2)} & {5.1+1.4} & {}\end{bmatrix}$

Solve

$H=\begin{bmatrix}{-0.48} & {-10.5} & {} \\ {5.1} & {-1.4} & {} \\ {-8} & {6.5} & {}\end{bmatrix}$

So, we find the element at address h31:

$H=\begin{bmatrix}{h11} & {h12} & {} \\ {h21} & {h22} & {} \\ {h31} & {h32} & {}\end{bmatrix}$

In this case, position h31 is - 8.0

8 0

1 year ago

Match each item in Column A to an answer in Column B.

aivan3 [116]

1. -> B. ;
2. -> C. ;
3. -> B. ;
4. -> A.. ;

7 0

3 years ago

Please help ! I have been stuck on this calculus problems, I will mark you brainliest!

Anna71 [15]

Answer:

$\displaystyle J'(3) = -1$

General Formulas and Concepts:

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative: $\displaystyle \frac{d}{dx} [e^u]=e^u \cdot u'$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle J(x) = e^{f(x)}$

Step 2: Differentiate

eˣ Derivative [Derivative Rule - Chain Rule]: $\displaystyle J'(x) = \frac{d}{dx}[e^{f(x)}] \cdot \frac{d}{dx}[f(x)]$
Simplify: $\displaystyle J'(x) = f'(x)e^{f(x)}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle J'(3) = f'(3)e^{f(3)}$
Substitute in function values: $\displaystyle J'(3) = -e^{0}$
Simplify: $\displaystyle J'(3) = -1$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

4 0

3 years ago