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lyudmila [28]
3 years ago
6

Which situation can be represented by -35 + 35? Choose all that apply.

Mathematics
2 answers:
guapka [62]3 years ago
8 0

Answer:

A diver is 35 feet below the surface of the water and then comes up to the surface.

siniylev [52]3 years ago
4 0

Answer: B & D

Step-by-step explanation:

In this scenario we want to come out with a situation where both sides are in an even playing field, as the equation solves to zero.

If a diver is 35 feet below water (-35) and he comes up to breathe, he goes up 35ft (35) meaning he is at the surface (0).

If Manuel is in debt for 35$ (-35) and then he pays off the 35$ (+35) then he is even (at 0).

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you and your family are making a trip to see a friend.you drive 56 miles per hour for the first 3 hours and then drive 63 miles
Aleonysh [2.5K]
56 x 3 = 168

63 x 5 = 315

315 + 168 = 483

483 miles
4 0
3 years ago
X to the power of three equals 343
Norma-Jean [14]
X^3= 343

Find the cube root of both sides
∛x^3 = ∛343

x= 7

Final answer: x=7
3 0
3 years ago
Help me this is due today
Arte-miy333 [17]

Answer:

There is no question bud

Step-by-step explanation:

Where is the question?

8 0
2 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
Woodland high school
lesantik [10]
That is correct because it’s a real high school
8 0
3 years ago
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