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2 years ago

I need help please!

Mathematics

1 answer:

o-na [289]2 years ago

5 0

Answer:0

Step-by-step explanation:

X = 0

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ASAP What is the sum of 16.87 + (–98.35)? –115.22 –81.48 81.48 115.22

lubasha [3.4K]

<h2>Answer:</h2><h3>-81.48</h3>

Solution,

16.87+(-98.35)

=16.87-98.35

= -81.48

Hope it helps

Good luck on your assignment

4 0

2 years ago

Read 2 more answers

Solve for linear inequality of x - 1 ≥ 4 + x

defon

$\huge\bf{ANSWER:}$

$\huge\mathrm{No\:solution.}\checkmark$

$\huge\bf{SOLUTION}:$

Hey there, hope you are having a wonderful day! :)

Let's solve this inequality.

First of all, let's move all the numbers to the right, using the $\bf{\underline{opposite\:\:operation\:}$ :

$\longmapsto$ $\sf{x-1\geq 4+x$

$\longmapsto\sf{x\geq 4+x+1$

$\longmapsto\sf{x\geq x+5$

Now, let's move the variables to the left:

$\longmapsto\sf{x-x\geq 5$

$\longmapsto\sf{0\geq 5$

Now, as you can see, we ended up with a false statement. 0 is NOT greater than 5.

Thus, there are no values of x that make the inequality true, and the inequality has

$\bold{No\:solutions.}$

Hope you find it helpful.

Feel free to ask if you have any doubts.

$\bf{-MistySparkles^**^*$

8 0

2 years ago

Over the course of a year, a company goes from 115 employees to 124 employees. Use the drop-down menus to build an equation that

erica [24]

9 would be the differance 124-115+1-1=9

3 0

3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

6 0

3 years ago

Read 2 more answers

I’m about to cry from frustration. Help please before I die

kvv77 [185]

Answer:

$\frac{1}{3}x+16=37$