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tangare [24]
3 years ago
14

Find the difference when the polynomial -5x^2+3x+8 is subtracted from the polynomial 2x^2+4x+1.

Mathematics
1 answer:
geniusboy [140]3 years ago
4 0

The difference is 7x^{2}  + 1x-7.

Step-by-step explanation:

Step 1:

The polynomial -5x^{2}+3x+8 is subtracted from the polynomial 2x^{2} + 4x +1.

If we write this as an equation, we get

(2x^{2} + 4x +1) -(-5x^{2}+3x+8).

To subtract the polynomials, we group up the terms based on their variables.

In this subtraction, there are two variables i.e. x^{2} and x and there is one constant term.

Step 2:

The subtraction of the x^{2} variables; 2x^{2}  -(-5x^{2} )  = 2x^{2} +5x^{2} = 7x^{2}.

The subtraction of the x variables; 4x - (3x) = 1x.

The subtraction of the constants; 1-(8) = -7.

So (2x^{2} + 4x +1) -(-5x^{2}+3x+8) = 7x^{2}  + 1x-7.

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Which of the numbers below belong to both the set of natural numbers and the set of integers?
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15  -  natural (1,2,3...15....) and integers (...-2,-1,0, 1,2,...15,...)
8 0
3 years ago
Simplify algebraic expression 9+5n-3-5n
nadezda [96]

Answer:

6

Step-by-step explanation:

You have to add like terms. So the 5n and -5n would cancel each other out (because 5-5=0) and 9-3=6.

3 0
4 years ago
I need help asap pls and thank you ;)
olga55 [171]

Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

3 0
3 years ago
Write the equation of a quadratic function whose graph has x-intercepts at x=3 and x=-4 and a y-intercept at 48.
jasenka [17]

Answer:

Step-by-step explanation:

okay i will answer but i dont see the question

5 0
3 years ago
Find the value of given expression<br><br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B44%20%2B%2077%7D%20" id="TexFormula
Naily [24]

Answer:

11

Step-by-step explanation:

square of 44 + 77

1st step = go with the number that is inside the square

that will 44+77 = 121

now using the square root , you may think about which number multiple by himself give you 121

11×11 =121

so square root of 121 = 11

8 0
3 years ago
Read 2 more answers
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