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3 years ago

I need help can yall help

Mathematics

2 answers:

Svetach [21]3 years ago

8 0

Answer:

360"

Step-by-step explanation:

Firstly, you need to be able to convert your feet to inches. Remember, one foot is equal to twelve inches. You multiply six inches by five feet and later multiply by twelve to convert to inches.

svp [43]3 years ago

4 0

I believe she can cut 360 6 inch strips from all of the rolls

Because:

the 5 rolls of ribbon each have six feet of ribbon on each roll, times the amount of inches in a foot

5(rolls of ribbon)*6(feet on each roll)*12(inches in a foot) = 360 (6 inch strips she can cut)

PLEASE give brainliest if correct!
have a nice day :-)

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4(x+3)= -24<br><br> solve for x

san4es73 [151]

See attached.

Hope this helps!

5 0

3 years ago

Your school's drama club charges $4 per student for admission to Our Town. The club borrowed $400 from parents to pay for costum

inna [77]

Answer:

125 students attend.

Step-by-step explanation:

We already know that the drama club made $500 (500 - 400 = 100)

divide 500 by 4 to find out how many students attend the play, and paid $4

500 ÷ 4 = 125

125 students attend.

6 0

3 years ago

An employee receives a weekly salary of $340 and a 6% commission on all sales.

Alina [70]

340+.06(sales)
340+.06(660)

$379.60

5 0

3 years ago

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

Solve the system.

gtnhenbr [62]

X = y + 11

2x + y = -2
2(y + 11) + y = -2
2y + 22 + y = -2
3y + 22 = -2
3y = -24
y = -8

x = y + 11
x = -8 + 11
x = 3

Solution set {3, -8} (C)

8 0

3 years ago