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2 years ago

What is the shape of 6x2 when it is graphed?

Mathematics

1 answer:

jonny [76]2 years ago

8 0

The expression 6x^2 will have the shape of a parabola when graphed

<h3>What are quadratic equations?</h3>

Quadratic equations are second-order polynomial equations and they have the form y = ax^2 + bx + c or y = a(x - h)^2 + k

<h3>How to determine the shape of 6x^2 when it is graphed?</h3>

The function expression is given as:

6x^2

Express the function as an equation.

So, we have

y = 6x^2

The above equation is a parabola or a quadratic equation.

All quadratic equations have the same shape and the shape is parabola

This means that the expression 6x^2 will have the shape of a parabola when graphed

Hence, the expression 6x^2 will have the shape of a parabola when graphed

Read more about quadratic equations at

brainly.com/question/1214333

#SPJ1

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How would 9/25 as a percent

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36% is the answer to the question

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3 years ago

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Factor completely 2x3 + 14x2 + 4x + 28.

Oxana [17]

6 + 28 + 4x +28

34 +28 + 4x

62 + 4x?

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3 0

3 years ago

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Gabriel wants to paint the model. How

enyata [817]

Answer:

<h2>The area of the base is 144 square inches.</h2><h2>The area of each triangular face is 66 square inches.</h2><h2>Grabiel needs 408 square inches of paint.</h2>

Step-by-step explanation:

The complete problem is attached.

Notice that the figure is a square pyramid, where its base dimensions are 12 inches by 12 inches, which represents an area of

$B=12 \times 12 = 144 \ in^{2}$

The slant height of the pyramid is 11 inches, which allow us to find the area of each triangle face

$A=\frac{1}{2}bh =\frac{1}{2}(12)(11)= 66 \ in^{2}$

But there are four triangle faces, so $A_{faces} =4(66)=264 \ in^{2}$ .

Therefore, the area of each triangular face is 66 square inches.

So, the total surface area would be the sum

$S=144+264=408 \ in^{2}$

Therefore, Gabriel needs 408 square inches to paint the whole model.

4 0

3 years ago

At the soup to nuts cafeteria, larry orders two pieces of toast and a bagel, which comes out to $\$1.30$. Curly orders a bagel a

kotegsom [21]

Let us take costs of a piece of toast = $t, , one bagel cost=$b and a muffin cost=$m.

Larry

Two pieces of toast and a bagel cost = $1.30

2t+b=1.30 -------------equation(1)

Let us solve the equation(1) for t in terms of b, because we need to find one bagel cost.

Subtracting b from both sides we get

2t+b-b=1.30-b

2t= 1.30-b

Dividing by 2 on both sides.

2t/2= (1.30-b)/2

t= (1.30-b)/2

Curly

A bagel and a muffin cost = $2.50.

b + m = 2.50 -------------equation(2)

Solving equation for m in terms of b, we get

m= 2.50-b.

Moe

A piece of toast, two bagels, and three muffins cost = $6.95

t + 2b + 3m = 6.95 ......................equation(3).

Substituting t= (1.30-b)/2 and m= 2.50-b in equation (3)

(1.30-b)/2 + 2b + 3(2.50-b) = 6.95 .

Multiplying each term by 2 to get rid 2 from denominator of (1.30-b).

2*(1.30-b)/2 + 2*2b + 2*3(2.50-b) = 2*6.95

1.30-b + 4b + 6(2.50-b) = 13.90.

1.30 - b + 4b +15 - 6b = 13.90

Combining like terms

-3b +16.30 = 13.90

Subtracting 16.30 from both sides.

-3b +16.30-16.30 = 13.90-16.30.

-3b= -2.4

Dividing both sides by -3.

-3b/-3 = -2.4/-3

b = 0.8

Therefore, cost of one bagel = $0.80.

7 0

3 years ago

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

4 years ago