All categories

3 years ago

What is the answer to (-9) - 16 =

Mathematics

1 answer:

Andrews [41]3 years ago

6 0

Answer: -25

Step-by-step explanation:

-9 - 16 = -25

-9-16 is the same as -9 + -16 which gives you -25

You might be interested in

PLEASE PLEASE HELP WILL MARK BRAINLY

Hitman42 [59]

Answer:

The second option (see attached image)

Step-by-step explanation:

You are looking for a box diagram that represents 9 units, and from those, clearly marked sections that contain 3/2 = 1.5 units.The idea is to count how many 1.5 units you have in 9 units.

The in the second diagram you see 9 boxes subdivided in half. Then outlined in red other smaller boxes of length 1.5 units. We can clearly see from the diagram that there are exactly 6 of these smaller 1.5 units red boxes to produce the total 9 unit object.

5 0

3 years ago

Help please i have a map test :)

just olya [345]

I think it could be E

4 0

3 years ago

Use the diagram to complete the statements.

andrey2020 [161]

Answer:

3, 4, Point R to Point Q, and depression from Point Q to Point R

Step-by-step explanation:

......

3 0

3 years ago

Read 2 more answers

A boy moves 4 miles south. Then he turns 90 degrees to the left. He moves forward 6 miles. H turns 90 degrees to the left. He mo

zaharov [31]

Y1=-4

x1=6

y2=4

y1+y2=0

So he is just his x displacement from where he started, 6 miles.

8 0

4 years ago

2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^

Ber [7]

I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

dy/dx = 2 exp(at)

Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

d²y/dx² = df/dt • dt/dx

d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

4 0

2 years ago