Evaluate the expression when x = 5 Am I correct? Is 5 the answer? I’m confused.

vesna_86 [32]

Answer:

verb tense refers to when the action and a sentence takes place - whether it happened in the past, is happening in the present, or what is happening in the future.

4 0

3 years ago

What is the first ordered pair and the second ordered pair?

Delvig [45]

When x=-5, y=(1/5)*(-5)-1=-2, so the first order pair is (-5,-2)
when y=-1, -1=(1/5)x-1, (1/5)x=0, x=0, so the second ordered pair is (0, -1)

7 0

3 years ago

If the exterior sides of two adjacent angles are opposite rays, then the angles are _____ angles.

elena-s [515]

The correct answer is B. Supplementary

That is the basic definition of what supplementary angles are and can be found in any textbook.

4 0

3 years ago

Read 2 more answers

You have 100 yards of fencing available to create an enclosure for farm animals. As the

Sunny_sXe [5.5K]

1. area of a pentagon when 1 side is x is $\frac{x^2}{4}\sqrt{5(5+2\sqrt{5})}$
so if perimiter is 100 then one side is 100/5 or 20
area will be $\frac{20^2}{4}\sqrt{5(5+2\sqrt{5})}$ ≈688.2 square yards

2. area of a hexagon when 1 side is length x is $\frac{3x^2\sqrt{3}}{2}$
so if permiter is 100 then length of one side is 100/6 or about 16.666666666666
the area will be $\frac{3(16.6666666)^2\sqrt{3}}{2}$ ≈721.7 square yards

3. area of a regular octagon with side lenghts x is $2(1+\sqrt{2})x^2$
so if 8 sides then side length of each is 100/8 or 12.5
area will be $2(1+\sqrt{2})12.5^2$ ≈754.4 square yards

octagon wil have most space
you can tell because the more sides it has, the more it gets to a circle and the more area it encloses with a given perimiter

3 0

3 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

4 0

3 years ago