What is the sum of the polynomials? (7x3 – 4x2) + (2x3

Dana wants to make two types of dog treats. She has 10 cups of peanut butter and 12 cups of flour. Her dog bone treat recipe use

alexdok [17]

D. Let x represent the number of trays of dog bone treats made and y represent number of trays of oatmeal dog treats made.

5 0

2 years ago

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The function g(x) = 5x2 – 10x written in vertex form is g(x) = 5(x – 1)2 – 5. The function g(x) is shown on the graph along with

natka813 [3]

The general vertex form of the a quadratic function is y = (x - h)^2 + k. In this form, the vertex is (h,k) and the axis of symmetry is x = h. Then, you only need to compare the vertex form of g(x) with the general vertex form of the parabole to conclude the vertex point and the axis of symmetry. g(x) = 5(x-1)^2 - 5 => h = 1 and k = - 5 => theis vertex = (1, -5), and the axis of symmetry is the straight line x = 1. Answer: the vertex is (1,-5) and the symmetry axis is x = 1.

8 0

2 years ago

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In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline

VLD [36.1K]

Answer:

The measure of the arc CD = 64°

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°

6 0

2 years ago

What type of polygon would a slice of a hexahedron at a vertex create? Explain. What type of polygon would a slice of an icosahe

harina [27]

Answer:

hexahedron: triangle or quadrilateral or pentagon
icosahedron: quadrilateral or pentagon

Step-by-step explanation:

Hexahedron

A hexahedron has 6 faces. A regular hexahedron is a cube. 3 square faces meet at each vertex.

If the hexahedron is not regular, depending on how those faces are arranged, a slice near a vertex may intersect 3, 4, or 5 faces. The first attachment shows 3- and 4-edges meeting at a vertex. If those two vertices were merged, then there would be 5 edges meeting at the vertex of the resulting pentagonal pyramid.

A slice near a vertex may create a triangle, quadrilateral, or pentagon.

Icosahedron

An icosahedron has 20 faces. The faces of a regular icosahedron are all equilateral triangles. 5 triangles meet at each vertex.

If the icosahedron is not regular, depending on how the faces are arranged, a slice near the vertex may intersect from 3 to 19 faces.

A slice near a vertex may create a polygon of 3 to 19 sides..

3 0

2 years ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0

2 years ago

What is the sum of the polynomials? (7x3 – 4x2) + (2x3 – 4x2)