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N76 [4]
3 years ago
13

5. What is the degree of the sum of these two polynomials? (5x3 - 2x2 + 1) + (2x2 - 6x4-5) A. 2 B.3 6.4 D.5​

Mathematics
1 answer:
hichkok12 [17]3 years ago
4 0

Answer:

14

Step-by-step explanation:

(5x3-2x2+1) + (2x2-6x4-5)

(15-2x2+1) + (4-6x4-5)

(13x2+1) + (-2x4-5)

(26+1) + (-8-5)

27 + -13

14

Hope this helps... I was confused on the question about the decimals this is the best I can do... Have a nice day though! ☺️

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PLEASE HELP, BEEN STUCK FOR AN HOUR
djverab [1.8K]

Answer:

Step-by-step explanation:

To sketch a quadratic function we need two things:

1)  Nature of the curve

2) Vertex

3) y-intercept

The completing square form of the quadratic equation is:

y=a(x-h)^2+k

where,

a represents the nature of the graph it can be maximum or minimum , meaning, if a > 0 then minimum(u shaped curve/happy face) and if a < 0 then maximum(n shaped curve/sad face).

h represents the x-coordinate of the vertex.

k represents the y-coordinate of the vertex.

Now if we compare g(x) with our completing square form we get the following:

g(x)=(x-4)^2+12\\y=a(x-h)^2+k\\

When we simply compare the following we get ,

a = 1 , which means a > 0 since 1 is greater than 0 the nature of the curve will be minimum(happy face/u shaped)

h = 4, which means the x-coordinate of the vertex is 4

k = 12, which means the y-coordinate of the vertex is 12

Now we have the nature of the curve, we have the vertex now all we need is the y-intercept.

For y-intercept:

For y-intercept meaning at which point will the graph cross the y-axis(0 , y)

For that we expand the formula and turn it into the standard quadratic equation form by using the formula (a - b)^2

y=(x-4)^2+12\\y=((x)^2-2(x)(4)+(4)^2)+12\\y=x^2-8x+16+12\\y=x^2-8x+28\\

now we compare with the standard quadratic form:

y=ax^2+bx+c

here c is the y-intercept and while comparing we can see that c = 28 ,

so the curve cuts the y-axis at (0 , 28)

So we have all the three things that we need to graph our function.

So we just plot the y-intercept , the vertex , and join the dots. Just a tip draw a dotted line on the x-coordinate of the vertex because the vertex point is also called as a turning point where the graph goes in the opposite direction just like a mirror reflection. I attached 2 images you can check them out. One is handmade(i know i suck at drawing but still xD) , one is sketched by online graphing calculator.

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nexus9112 [7]

Answer: The zeros are x= -1 and x= -3

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Answer:

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Step-by-step explanation:

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What is 851 divided by 48?
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Answer:

= 17.7 ( rounded off to one decimal place)

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Step-by-step explanation:

851 ÷ 48

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Read 2 more answers
What is the largest possible integral value in the domain of the real-valued function
kotegsom [21]

Answer:

Max Value: x = 400

General Formulas and Concepts:

<u>Algebra I</u>

  • Domain is the set of x-values that can be inputted into function f(x)

<u>Calculus</u>

  • Antiderivatives
  • Integral Property: \int {cf(x)} \, dx = c\int {f(x)} \, dx
  • Integration Method: U-Substitution
  • [Integration] Reverse Power Rule: \int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = \frac{1}{\sqrt{800-2x} }

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

u = 800-2x\\du = -2dx

<u>Step 3: Integrate</u>

  1. Define:                                                                                                            \int {f(x)} \, dx
  2. Substitute:                                                                                         \int {\frac{1}{\sqrt{800-2x} } } \, dx
  3. [Integral] Int Property:                                                                                     -\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx
  4. [Integral] U-Sub:                                                                                           -\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du
  5. [Integral] Rewrite:                                                                                          -\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du
  6. [Integral - Evaluate] Reverse Power Rule:                                                 -\frac{1}{2}(2\sqrt{u}) + C
  7. Simplify:                                                                                                         -\sqrt{u} + C
  8. Back-Substitute:                                                                                            -\sqrt{800-2x} + C
  9. Factor:                                                                                                           -\sqrt{-2(x - 400)} + C

<u>Step 4: Identify Domain</u>

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

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3 years ago
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