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saw5 [17]
2 years ago
6

Covert each into algebraic equations

Mathematics
1 answer:
marshall27 [118]2 years ago
3 0

Answer:

1) 2x=4x-8

2) 4(x+2)=20

3) 2x-9=4(x+5)

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ILL GIVE BRAINLIES!!!!<br> Please explain! Thank you!
givi [52]

Answer:

I think A but sorry i might be wrong

Step-by-step explanation:

8 0
2 years ago
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Solve for x:<br> 2x + y = -2<br> 4x + y = -8
bagirrra123 [75]

Answer:

x = -3

Step-by-step explanation:

2x + y = -2

4x + y = -8

2x + y = - 2

y = -2x - 2

4x + y = - 8

y = -4x - 8

y = -2x - 2

y = -4x - 8

-2x - 2 = -4x - 8

4x - 2x = 2 - 8

2x = -6

2x/2 = -6/2

x = -3

To check just use da graph ;w;

The x is -3

7 0
3 years ago
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Multiply. Write in simplest form. Identify the two whole numbers between which the product lies 24 * 1/5
lutik1710 [3]
24*(1/5)=4.8
the answer would then be 4 and 5
5 0
3 years ago
Can someone please help??? WILL MARK BRAINLIEST
olya-2409 [2.1K]

y = mx + b

"m" is the slope, "b" is the y-intercept (the y value when x = 0)


To find the slope, you can use the slope formula and plug in the two points:

(3,63) = (x₁ , y₁)

(5,107) = (x₂, y₂)

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m=\frac{107-63}{5-3}

m=\frac{44}{2}

m = 22


Now that you know m = 22, plug it into the equation:

y = mx + b

y = 22x + b

To find "b", plug in one of the points into the equation (I will do both points)

(3,63)

y = 22x + b

63 = 22(3) + b

63 = 66 + b    Subtract 66 on both sides

-3 = b


(5,107)

y = 22x + b

107 = 22(5) + b

107 = 110 + b    Subtract 110 on both sides

-3 = b


y = 22x - 3

8 0
3 years ago
(1.1/1.2: Interpolating polynomials) Say we want to find a polynomialf(x) ofdegree 3,f(x) =a0+a1x+a2x2+a3x3,satisfying some inte
Hunter-Best [27]

(a) If

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then from the given conditions we get the system of equations,

<em>f</em> (-1) = <em>a</em>₀ - <em>a</em>₁<em> </em>+<em> a</em>₂ - <em>a</em>₃ = -1

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+ 4<em>a</em>₂ + 8<em>a</em>₃ = 1

<em>f</em> (3) = <em>a</em>₀ + 3<em>a</em>₁<em> </em>+<em> </em>9<em>a</em>₂ + 27<em>x</em> ³ = 5

(b) Similarly, if

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then

<em>f'(x)</em> = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ <em>x</em> + 3<em>a</em>₃ <em>x</em> ²

so that the given conditions yield the system,

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 0

<em>f'</em> (1) = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ + 3<em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 27<em>a</em>₃ = 3

<em>f'</em> (2) = <em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 12<em>a</em>₃ = -1

7 0
2 years ago
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