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1 year ago

What -7 x4=36? amogus amogus amogus amogus

Mathematics

1 answer:

77julia77 [94]1 year ago

3 0

Based on the given problem of -7x⁴ = 36, the solution to the equation is x = -1.5.

<h3>What is the solution to this problem?</h3>

The equation to find x is given as:

-7x⁴ = 36

Solving it gives:

-7x⁴ = 36

x⁴ = 36/-7

x⁴ = -36/7

x = ⁴√(-36/7)

x = -1.5

In conclusion, x is -1.5.

Find out more questions on solving for x at brainly.com/question/2910769

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Orlov [11]

The answer would be:

A translation 5 units down, followed by a 180-degree counterclockwise rotation about the origin

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2 years ago

Which equation represents a line that passes through the point ( 6 , − 3 ) and is parallel to the graph of y = 3 x + 1 ?

Ratling [72]

Answer:

y + 3= 3(x - 6) ← point-slope form

y = 3x - 21 ← slope-intercept form

3x - y = 21 ← standard form

Step-by-step explanation:

The point-slope form of the equation of line it's y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point the line passing through.

y=m₁x+b₁ ║ y=m₂x+b₂ ⇔ m₁ = m₂

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y = 3x + 1 ⇒ m₁ = 3 ⇒ m₂ = 3

(6, -3) ⇒ x₁ = 6, y₁ = -3

point-slope form:

y - (-3) = 3(x - 6)

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2 years ago

Which value is an input of the function? –14 –2 0 4

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Step-by-step explanation:

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3 years ago

Given: JK tangent, KH=16, HE=12 Find: JK.

Y_Kistochka [10]

Answer: $JK=8$

Step-by-step explanation:

You can observe in the figure that JK is a tangent and KH is a secant and both intersect at the point K. Then, according to the Intersecting secant-tangent Theorem:

$JK^2=KE*KH$

You know that:

$KH=KE+HE$

Then KE is:

$KE=KH-HE$

$KE=16-12$

$KE=4$

Now you can substitute the value of KE and the value of KH into $JK^2=KE*KH$ and solve for JK. Then the result is:

$JK^2=4*16\\JK^2=64\\JK=\sqrt{64}\\JK=8$

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2 years ago

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max2010maxim [7]

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3 years ago

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