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rosijanka [135]
3 years ago
13

Which of the following equations are equivalent to -2m - 5m - 8 = 3 + (-7) + m?

Mathematics
2 answers:
Anika [276]3 years ago
8 0
The answer is 2 ...............

12345 [234]3 years ago
8 0
3............;::::::::::::::
You might be interested in
1 point
Anton [14]

Answer:

  C.  350 miles

Step-by-step explanation:

Daphne wants to drive 3/4 of the distance in the first 2 days. So, her mileage per day should be about (3/4)/2 = 3/8 of the total mileage:

  (3/8)(932.4 mi) = 349.65 mi ≈ 350 mi

She should drive about 350 miles on each of the first 2 days.

8 0
3 years ago
Who can help and explain this plzz
MariettaO [177]
They could have travelled any distance. In total, they travelled 1.75 times the distance of the first day, but since we do not know the distance of the first day, we have no idea the actual distance. For all we know they could have gone to the moon in the first day. 
6 0
3 years ago
Compute the Taylor expansion of order n=2 of the function sin(xy) at x=0 and y=0
artcher [175]

Answer:

f(x, y) = Sin(x*y)

We want the second order taylor expansion around x = 0, y = 0.

This will be:

f(x,y) = f(0,0) + \frac{df(0,0)}{dx} x + \frac{df(0,0)}{dy} y + \frac{1}{2} \frac{d^2f(0,0)}{dx^2} x^2 +\frac{1}{2} \frac{d^2f(0,0)}{dy^2}y^2  + \frac{d^2f(0,0)}{dydx} x*y

So let's find all the terms:

Remember that:

\frac{dsin(ax)}{dx}  = a*cos(ax)

\frac{dcos(ax)}{dx} = -a*cos(ax)

f(0,0) = sin(0*0) = 1.

\frac{df(0,0)}{dx}*x = y*cos(0*0)*x = x*y

\frac{df(0,0)}{dy} *y = x*cos(00)*y = x*y

\frac{1}{2} \frac{d^2f(0,0)}{dx^2}*x^2 =  -\frac{1}{2}  *y^2*sin(0*0)*x^2 = 0

\frac{1}{2} \frac{d^2f(0,0)}{dy^2}*y^2 =  -\frac{1}{2}  *x^2*sin(0*0)*y^2 = 0

\frac{d^2f(0,0)}{dxdy} x*y = (cos(0*0) -x*y*sin(0*0))*x*y = x*y

Then we have that the taylor expansion of second order around x = 0 and y = 0 is:

sin(x,y) = x*y + x*y + x*y = 3*x*y

6 0
2 years ago
A math textbook might weigh _______________ .
djyliett [7]

Answer:

Depending on the book, B, 25 ounces!

Step-by-step explanation:

Please mark as brainliest

5 0
2 years ago
Read 2 more answers
n ΔABC, AB = 10 and BC = 5. Which expression is always true? A. 5 < AC < 10 B. AC = 5 C. 5 < AC < 15 D. AC = 10
ollegr [7]

Answer:

A. 5 < AC < 10

Step-by-step explanation:

We are given the ∆ABC

To solve for the sides of ∆ABC , we make use of Pythagoras Theorem

Pythagoras Theorem states that:

AB² = AC² + BC²

Where AB = Longest side, hence its length is always more that AC and BC.

In the above Question, we are given the values of AB and BC

AB = 10

BC = 5

Inputting these values into the Pythagoras Theorem,

10² = AC² + 5²

100 = AC² + 25

AC² = 100 - 25

AC² = 75

AC = √75

AC = 8.6602540378

According to the above calculation, we can see that the expression that is always true = Option A "5 < AC < 10"

4 0
3 years ago
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