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adoni [48]
3 years ago
12

What is the value of x in the equation log5 x = 4 jmap?

Mathematics
1 answer:
steposvetlana [31]3 years ago
5 0
To solve for x we proceed as follows:
from the laws of logarithm, given that:
log_a b=c
then
a^c=b
applying the rationale to our question we shall have:
log_5 x=4
hence
5^4=x
x=625
Answer: x=625
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Allison went to the store to buy some walnuts. The price per pound of the walnuts is
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Step-by-step explanation:

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A painting crew reported that a job is 3/5 completed. What fraction of the job remains to be done?
Lunna [17]

Answer:

2/5

Step-by-step explanation:

5/5 = full job

3/5  = done

full job - done =  remaining

5/5 job - 3/5 job

= 2/5 (remaining)

3 0
3 years ago
Read 2 more answers
Solve the equation. -3x+2x=8​
Andre45 [30]

Answer:

= − 8

Step-by-step explanation:

Combine like terms  

Divide both sides of the equation by the same term

Simplify

8 0
3 years ago
Which statement is true about ∠1 and ∠2?
gavmur [86]

Answer: Most likely A

Step-by-step explanation:

the reason I say its A is because of the face that congruent means in line not and straight. Though in your answer it shows that they are not alined so that leaves us with complemantary and the 2 complementary answer choices are

A & C but C is wrong because it say alternate which also means the that they would be 2 different angels so yeah I'm gonna go for A.

Good luck :)

5 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
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