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Olegator [25]
2 years ago
6

Find the degrees and terms6x⁴ + 3x² + 9 =​

Mathematics
1 answer:
Ivenika [448]2 years ago
3 0

Answer:

3 Terms and Degree is 6

Step-by-step explanation:

3 Terms and Degree is 6

How to find degrees:

4 + 2 = 6

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Elis [28]

Answer:

(-6,5)

Step-by-step explanation:

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shusha [124]

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Step-by-step explanation:

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8 0
3 years ago
A tank pumps out liquid at a rate of 12 gallons every 20 days. What is the unit rate in gallons per week?
pashok25 [27]

Answer:

4.2

Step-by-step explanation:

simplify 12 and 20 to 6 and 10 then divide 6 by 10 and multiply by 7 to get the answer.

8 0
3 years ago
Which is equivalent to (2^-3)^4 <br><br> 1/2^12<br><br> 2<br><br> -2^12<br><br> 1/2^7
sleet_krkn [62]

Answer:

The answer is First option

1/2^12 .

\frac{1}{2^{12} }

Step-by-step explanation:

Given:

(2^-3)^4

i.e

(2^{-3} )^{4}

To Find:

(2^{-3} )^{4}=?

Solution:

Law of Indices

(x^{a}) ^{b} =x^{a\times b}\\and\\x^{-1}=\frac{1}{x}

Using the above Property we get

∴ (2^{-3} )^{4}=2^{-3\times 4} \\\\ (2^{-3} )^{4}=2^{-12} \\\\(2^{-3} )^{4}= \frac{1}{2^{12} }

The answer is First option

1/2^12 .

(2^{-3} )^{4}=\frac{1}{2^{12} }

6 0
4 years ago
Eliminate the parameter and obtain the standard form of the rectangular equation. line through (x1, y1) and (x2, y2): x = x1 + t
Helen [10]

The parametric equations for the line passes through the points (X₁ , Y₁) and (X₂ , Y₂) are :

X = X₁ + t (X₂ - X₁) ......................... (1)

Y = Y₁ + t (Y₂ - Y₁) .......................... (2)

For eliminating the parameter (t) , first we will solve any one equation for 't' and then substitute that into another equation.

From the equation (1),

⇒ t = \frac{X- X1}{(X2 - X1)}

Substituting this t = \frac{X- X1}{(X2 - X1)} into the equation (2), we will get :

Y = Y₁ + \frac{(X- X1)}{(X2 - X1)} (Y₂ - Y₁)

Y = Y₁ + \frac{(X- X1)(Y2 - Y1)}{(X2 - X1)}

So, this is the standard form of rectangular equation.

For two given points (0, 0) and (4, -4)

X₁ = 0 , Y₁ = 0, X₂ = 4 and Y₂ = -4

For finding the parametric equations, we will plug these values into the given parametric equations.

X = X₁ + t (X₂ - X₁)

⇒ X = 0+ t (4 - 0)

⇒ X = 4t

and Y = Y₁ + t (Y₂ - Y₁)

⇒ Y = 0+ t (-4 - 0)

⇒ Y = - 4t

So, the parametric equations for the line passing through (0,0) and (4, -4) are: X = 4t and Y = -4t

6 0
3 years ago
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