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2 years ago

Solve the equation. m + (-22) = -25

Mathematics

1 answer:

lutik1710 [3]2 years ago

5 0

Answer:

m=-3

Step-by-step explanation:

move the -22 to the other side by adding 22 on both sides:

m-22+22=-25+22

Simplify:

m=-3

Answer: m=-3

Hope this helps!

You might be interested in

Give the sum in simplest form: a/c + b/c = _____

djverab [1.8K]

Answer:

(a+b)/c

Step-by-step explanation:

a/c+b/c

since they have the same denominator you can add the numerator together

(a+b)/c

6 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

2 years ago

If a position on the ocean floor is 12 kilometers below the sea level, which value describes the elevation of the ocean floor

LenKa [72]

Answer:

- 12

Step-by-step explanation:

The sea level serves as a reference point and may be used to describe objacetsvir positions above or below it. Therefore, the sea level is regarded as a zero point or level. Hence, objects below are describes to be at a negative distance while those above are assigned the positive sign. Hence, for a position which is 12 km below sea level, then the object is said to be at a distance of - 12 km (below sea level takes the negative sign.)

4 0

2 years ago

A system of two equations is shown below. What will you need to multiply the

irga5000 [103]

Answer:

the right answer is : -3

Step-by-step explanation:

$\left \{ {{x+y=4} \atop {3x+4y=14}} \right. \\\left \{ {{-3(x+y)=-3*4} \atop {3x+4y=14}} \right. \\\left \{ {{-3x-3y=-12} \atop {3x+4y=14}} \right. \\\left \{ {{-3x-3y+3x+4y=2} \atop {3x+4y=14}} \right. \\\left \{ {{-3y+4y=2} \atop {3x+4y=14}} \right. \\\left \{ {{y=2} \atop {3x+4*2=14}} \right. \\\left \{ {{y=2} \atop {3x=14-8=6}} \right. \\\left \{ {{y=2} \atop {x=\frac{6}{3}=2 }} \right.$

3 0

2 years ago

What are the x and y-intercepts of the line described by the equation?

pishuonlain [190]

To find the x-intercept, substitute 0 for y and solve for x.
-6x + 3(0) = 18.9
-6x = 18.9
x = 18.9 / -6
x = -3.15 <-- Your x-intercept is (-3.15,0)

To find the y-intercept, substitute in 0 for x and solve for y.
-6(0) + 3y = 18.9
3y = 18.9
y = 18.9/3
y = 6.3 (0,6.3) <-- Your x-intercept is (0,6.3)

If this helped, please mark me brainliest, thank you and if you need more help, feel free to message me!

8 0

2 years ago