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3 years ago

Will mark brainiest! Select all that apply.

Mathematics

1 answer:

Gala2k [10]3 years ago

4 0

I don’t see the pic so I can’t help

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Write an inequality for x that would give this rectangle a perimeter of at least 300ft

Travka [436]

Answer:

The inequality for $x$ is:

$x\geq 142$

Step-by-step explanation:

Given:

Width of rectangle = 3 ft

Height or length of rectangle = $(x+5)$ ft

Perimeter is at least 300 ft

To write an inequality for $x$ .

Solution:

Perimeter of a rectangle is given as:

⇒ $2l+2w$

where $l$ represents length of the rectangle and $w$ represents the width of the rectangle.

Plugging in the given values in the formula, the perimeter can be given as:

⇒ $2(x+5)+2(3)$

Using distribution:

⇒ $2x+10+6$

Simplifying.

⇒ $2x+16$

The perimeter is at lest 300 ft. So, the inequality can be given as:

⇒ $2x+16\geq 300$

Solving for $x$

Subtracting both sides by 16.

⇒ $2x+16-16\geq 300-16$

⇒ $2x\geq284$

Dividing both sides by 2.

⇒ $\frac{2x}{2}\geq \frac{284}{2}$

⇒ $x\geq 142$ (Answer)

7 0

4 years ago

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating

MakcuM [25]

Answer:

A=8.4063 $u^{2}$

Step-by-step explanation:

Be the functions:

$y=\frac{3}{x};y=\frac{3}{x^{2}}:x=7$

according the graph:

$\int\limits^1_7 {\frac{3}{x} } \, dx -\int\limits^1_7 {\frac{3}{x^{2} } } \, dx =3\int\limits^1_7 {\frac{1}{x} } \, dx -3\int\limits^1_7 {\frac{1}{x^{2} } } \, dx=3(\int\limits^1_7 {\frac{1}{x} } \, dx -\int\limits^1_7 {\frac{1}{x^{2} } } \, dx)=3[lnx-\frac{1}{x}](1-7)=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}$