All categories

3 years ago

Mine A is 225 feet underground and Mine B has an elevation of -223 feet. Which mine is further underground

Mathematics

2 answers:

Vitek1552 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

Mine A, Mine B is not even a mine, its negative, not a hole

LUCKY_DIMON [66]3 years ago

3 0

Answer: Mine A

Step-by-step explanation: Mine A is further underground because elevation means height and when it’s shows a number that is negative that means underground. 225 is greater than 223. So Mine A is further underground.

You might be interested in

I do not understand this at all. help please

Len [333]

Equation:

-9x+10y=-9

10y=9x-9

y=9/10x-9/10

Slope: 9/10

Y intercept: -9/10

So your answer is D

5 0

3 years ago

Read 2 more answers

Combine the like terms to create an equivalent expression:2r+1+(−4r)+7

STatiana [176]

Answer:

2r+1+(-4r)+7

= -2r+8

= -2(r+4)

Step-by-step explanation:

3 0

2 years ago

2. CAN SOMEONE PLEASE HELP ME? I'M NOT GOOD IN MATH. Explain your work

Elodia [21]

Answer:

Step-by-step explanation:

Since triangles ABC and CDB are similar, CD and CB is the same as the ratio between CB and CA. 15:9=x:15. x=25. Subtracting CD to just be left with AD, we get 25-9=16, or answer choice D. Hope this helps!

6 0

3 years ago

Alaine has 1 gallon of paint. She is going to pour it into a paint tray that measures 10 inches wide, 12 inches long, and 5 cm d

Alona [7]

She is going to pour it into a paint tray that measures 10 inches wide, 12 inches long, and5 cm deep. (1 gallon =... ... (1 gallon = 231 in3, 1 inch = 2.54 cm) Which of the following scenarios willhappen? a. The paint will not fill it

8 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side so proof is complete

8 0

3 years ago

Read 2 more answers