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2 years ago

Write the direct variation function given that y varies directly with x, and y = 1.5 when x = 5.

Mathematics

1 answer:

Cloud [144]2 years ago

4 0

Answer:

y=0.3x

Step-by-step explanation:

Formula = $K=\frac{y}{x}$

y α x

y = kx

Where k is the constant of proportionality.

When y = 1.5; x = 5;

We substitute these known values in the equation,

y = kx

1.5 = k5

Dividing both sides of the equation by 5 to find the value for k, we have

1.5/5= k0.3/5

Therefore,

K = 0.3

Having found the value of k,

We substitute this value into the relationship

y = kx

Therefore we have,

y = 0.3x.

The direct variation function is therefore,

y = 0.3x.

[RevyBreeze]

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Step-by-step explanation:

The student made a mistake by identifying the maximum point by the x coordinate value of the vertex. Minimum or maximum points are determine using the y coordinate value.The student can determine the difference between a maximum and minimum by identifying the y coordinate of the vertex.

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3 years ago

X=-3y-5 3x+9y=7 how can u solve this?

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Answer:

Step-by-step explanation:

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3 years ago

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prohojiy [21]

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Step-by-step explanation:

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3 years ago

Use the properties of quadrilaterals to complete the statements. A has exactly one pair of parallel sides. A must have two pairs

vesna_86 [32]

Answer:

1. Trapezoid

2. Kite

3. Square, Rhombus

Step-by-step explanation:

Given - Use the properties of quadrilaterals to complete the statements.

To find - A ..... has exactly one pair of parallel sides.

A ....... must have two pairs of adjacent sides congruent.

A ........ must have all congruent sides.

Proof -

A Trapezoid has exactly one pair of parallel sides.

A Kite must have two pairs of adjacent sides congruent.

A Square, Rhombus must have all congruent sides.

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A trapezoid is a quadrilateral with exactly one pair of parallel sides.

A kite is a quadrilateral with exactly two pairs of adjacent congruent sides.

A rhombus is a parallelogram with four congruent sides.

A square can be defined as a rhombus which is also a rectangle . In other words, a square is a parallelogram with four congruent sides and four right angles.

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3 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ , cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$

We know that $6\sqrt{5+\sqrt{5}} = 16.13996\dots$ and $-\:6\sqrt{3-\sqrt{5}} = -5.24419\dots$ . Therefore,

Solution : $16.13996 - 5.24419i$

Which rounds to about option b.

7 0

3 years ago