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Semmy [17]
3 years ago
9

What is the answer for 1+1+7x​2​​+7y​3​​+3+2x​2​​−y​3​​

Mathematics
1 answer:
natta225 [31]3 years ago
6 0
First, a little housekeeping!!

1+1+7x​2​​+7y​3​​+3+2x​2​​−y​3​​ needs to be written as <span>1+1+7x​^2​​+7y​^3​​+3+2x​^2​​−y​^3.  The " ^ " signifies exponentiation and is essential here.

Now combine like terms:

7y^3-y^3   +   7x^2 + 2x^2 + 1 + 1 + 3

You'll get     6y^3 +9x^2 +  5.​​</span>
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Write each fraction as a percent.<br> 14/40
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35%

Step-by-step explanation:

Multiply each side by 2.5 to get 35/100 which will give you 35%.

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Solve for x,y,z on the triangle
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X = 64.5
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What is the answer to -3(1+6r)=14-r
Ilia_Sergeevich [38]
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3 years ago
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Let f(x) = 7x^2-5x+3 and g(x) = 2x^2+4x-6
Troyanec [42]

Answer:

\large\boxed{A.\ f(x)+g(x)=9x^2-x-3}\\\boxed{B.\ f(x)-g(x)=5x^2-9x+9}\\\boxed{g(x)-f(x)=-5x^2+9x-9}

Step-by-step explanation:

f(x)=7x^2-5x+3,\ g(x)=2x^2+4x-6\\\\A:\\f(x)+g(x)=(7x^2-5x+3)+(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3+2x^2+4x-6\qquad\text{combine like terms}\\f(x)+g(x)=(7x^2+2x^2)+(-5x+4x)+(3-6)\\f(x)+g(x)=9x^2-x-3

B:\\f(x)-g(x)=(7x^2-5x+3)-(2x^2+4x-6)\\f(x)+g(x)=7x^2-5x+3-2x^2-4x+6\qquad\text{combine like terms}\\f(x)-g(x)=(7x^2-2x^2)+(-5x-4x)+(3+6)\\f(x)+g(x)=5x^2-9x+9

C:\\g(x)-f(x)=(2x^2+4x-6)-(7x^2-5x+3)\\g(x)-f(x)=2x^2+4x-6-7x^2+5x-3\qquad\text{combine like terms}\\g(x)-f(x)=(2x^2-7x^2)+(4x+5x)+(-6-3)\\g(x)-f(x)=-5x^2+9x-9

3 0
3 years ago
How many real roots would f(x)=<br> x^3+2x^2+2x-5 have
goldfiish [28.3K]

Answer:

One real root

Step-by-step explanation:

By the fundamental theorem of algebra, an nth degree polynomial has n-possible real roots.

If there are complex roots, then the the complex roots come in pairs.

Therefore the number of possible real roots of f(x)=x^3+2x^2+2x-5 are 3 real roots with no complex pairs or 1 real root with a complex pair.

By Descartes rule of signs, there is only one change of sign in the polynomial (+ to -).

Hence there is only one positive real root.

f(-x)=-x^3+2x^2-2x-5

There is change is sign two times but we can not have even number of real roots for this polynomial

Therefore there is only real root.

7 0
4 years ago
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