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2 years ago

10

What equation could be written for this table?

1 answer:

kkurt [141]2 years ago

7 0

Answer:

y = 3 + 1/2 * x

Step-by-step explanation:

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A rectangular school banner has a length of 60 inches and a width of 48 inches. A sign is made that is similar to the school ban

natulia [17]

Answer:

The ratio of the length of the banner to the length of the sign = 54/17

So....since they are similiar figures, the widths will have the same ratio

And the ratio of the area of the banner to the area of the sign will be the square of this ratio

Step-by-step explanation:

[54/17]^2 = 2916/289

3 0

3 years ago

Can someone pls help me pls

In-s [12.5K]

Answer:

Answer attached

Step-by-step explanation:

7 0

2 years ago

*BRAINIEST & 20 POINTS*<br><br> !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

kondor19780726 [428]

Answer:

8 = x

y = 29

Step-by-step explanation:

4x + 17 = y + 20 because vertical angles theorem.

y + 20 = 5x + 9 because interior angles theorem.

4x + 17 = 5x + 9 because same side interior angles theorem.

Subtract 9 from both sides and subtract 4x from both sides.

8 = x

Plug 8 in for x and solve.

5 times 8 is 40, plus 9 equals 49. Subtract 20 from both sides and you get 29.

5(8) + 9 = y + 20

49 = y + 20

29 = y

I hope this helps!

#teamtrees #WAP (Water And Plant)

5 0

2 years ago

Read 2 more answers

PLEASE HELP!!! The ratio of footballs to soccer balls at a sporting goods store is 5 to 3. if the store has 100 footballs in sto

xxMikexx [17]

Answer:

70 or 60 is the answer

8 0

3 years ago

Read 2 more answers

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago