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chubhunter [2.5K]
3 years ago
12

\begin{aligned} &A = 7x^2-3x+10 \\\\ &B = -4x^2+6x-4 \end{aligned} ​ A=7x 2 −3x+10 B=−4x 2 +6x−4 ​ A-B=A−B=A, minus, B,

equals Your answer should be a polynomial in standard form.
Mathematics
1 answer:
fgiga [73]3 years ago
4 0

Answer:

<h2>A-B = 11x²-9x+14</h2>

Step-by-step explanation:

Given the polynomial functions A = 7x^2-3x+10 and B = -4x^2+6x-4, we are to find the value of A - B.

A-B =  7x²-3x+10 - ( -4x²+6x-4)

open the parenthesis

A-B = 7x²-3x+10 + 4x²-6x+4

collect the like terms

A-B  = 7x² + 4x²-3x-6x+10+4

A-B = 11x²-9x+14

<em>Hence the difference between both quadratic equation in standard form is 11x²-9x+14</em>

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Stefan sells Jin a bicycle for $147 and a helmet for $18. The total cost for Jim is 150% of what Stefan spent originally to buy
soldi70 [24.7K]

Answer:

He spent $110.00 on both originally

Step-by-step explanation:

Add both prices too get 165

Then use the equation \frac{150}{100} = \frac{165}{x}

cross multiply to get 150x=16500

Divide 16500 by 150 to get x alone

x= 110

4 0
3 years ago
Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

It is a homogeneous solution:

y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
The box plot below describes a data set. Which of the following has a value of 18?
Aloiza [94]
What box plot are you talking about
8 0
3 years ago
Read 2 more answers
Can someone please help me with this?
patriot [66]
The answer is 3y=-9x-6
7 0
3 years ago
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Answer:

D is correct

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Step-by-step explanation:

8 0
3 years ago
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