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4 years ago

Please answer my question now in two minutes

Mathematics

1 answer:

Fudgin [204]4 years ago

5 0

$\displaystyle\bf\\UX=\frac{VW}{2}\\\\\implies~2UX=VW\\\\ 2z=z+34\\2z-z=34\\z=34\\\\\implies~\boxed{\bf UX=34}$

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viktelen [127]

Answer:

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3 years ago

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The Quan’s mother made a quilt for his bed. The quilt is 2.44 meters long And 1.83 meters wide. What is the area of the quilt in

Katarina [22]

Your answer is 48.11 sqft.

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3 years ago

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What is measure of side A?

Ira Lisetskai [31]

<u>Answer:</u>

The correct answer option is c. a = 11.94.

<u>Step-by-step explanation:</u>

We are given a right angles triangle with an angle ABC equal to 58 degrees and side length AC known to be equal to 19.1.

We are supposed to find the length of side a.

Side AC, opposite to angle ABC, will be the perpendicular and BC will be the base so we will use the formula of tan to find b.

$tan 58 = \frac{19.1}{a}$

$a = \frac{19.1}{tan 58}$

$a=11.94$

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4 years ago

Nicholas wants to buy a new computer that is on sale for $899. His parents will loan him the money if he repays them. He has $22

Afina-wow [57]

Answer:

20 days

Step-by-step explanation:

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3 years ago

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Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.

Levart [38]

Answer:

The standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Step-by-step explanation:

We know that:

$4p\left(y-k\right)=\left(x-h\right)^2$ is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

$x^2\:+\:10x\:+\:6y\:=\:47$

Rewriting the equation in the standard form

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by $x^2\:+\:10x\:+\:6y\:=\:47$ is:

$4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2$

where

vertex (h, k) = (-5, 12)

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3 years ago