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2 years ago

What is n -8+4(1+5n)=-6n-14

Mathematics

1 answer:

jeka942 years ago

4 0

Answer:

n = (-5)/13

Step-by-step explanation:

Solve for n:

4 (5 n + 1) - 8 = -6 n - 14

4 (5 n + 1) = 20 n + 4:

20 n + 4 - 8 = -6 n - 14

Add like terms. 4 - 8 = -4:

20 n - 4 = -6 n - 14

Add 6 n to both sides:

20 n + 6 n - 4 = (6 n - 6 n) - 14

6 n - 6 n = 0:

20 n + 6 n - 4 = -14

20 n + 6 n = 26 n:

26 n - 4 = -14

Add 4 to both sides:

26 n + (4 - 4) = 4 - 14

4 - 4 = 0:

26 n = 4 - 14

4 - 14 = -10:

26 n = -10

Divide both sides of 26 n = -10 by 26:

(26 n)/26 = (-10)/26

26/26 = 1:

n = (-10)/26

The gcd of -10 and 26 is 2, so (-10)/26 = (2 (-5))/(2×13) = 2/2×(-5)/13 = (-5)/13:

Answer: n = (-5)/13

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Tanya [424]

Let's solve your system by substitution.

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2 years ago

Mrs. Gilbert is interested in determining how many students spend 4 or more hours doing homework each week. She asks every stude

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Answer:

C. Biased, because she only asked students on honor roll.

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Since Mrs. Gilbert's goal was to find how many students that spent >4 hours and not how many students that earned Honor Roll that spent >4 hours, the data is biased. Choose C.

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2 years ago

Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) ∈ R if and only if ad = bc. Ar

Veseljchak [2.6K]

Answer:

The given relation R is equivalence relation.

Step-by-step explanation:

Given that:

$((a, b), (c, d))\in R$

Where $R$ is the relation on the set of ordered pairs of positive integers.

To prove, a relation R to be equivalence relation we need to prove that the relation is reflexive, symmetric and transitive.

1. First of all, let us check reflexive property:

Reflexive property means:

$\forall a \in A \Rightarrow (a,a) \in R$

Here we need to prove:

$\forall (a, b) \in A \Rightarrow ((a,b), (a,b)) \in R$

As per the given relation:

$((a,b), (a,b) ) \Rightarrow ab =ab$ which is true.

$\therefore$ R is reflexive.

2. Now, let us check symmetric property:

Symmetric property means:

$\forall \{a,b\} \in A\ if\ (a,b) \in R \Rightarrow (b,a) \in R$

Here we need to prove:

$\forall {(a, b),(c,d)} \in A \ if\ ((a,b),(c,d)) \in R \Rightarrow ((c,d),(a,b)) \in R$

As per the given relation:

$((a,b),(c,d)) \in R$ means $ad = bc$

$((c,d),(a,b)) \in R$ means $cb = da\ or\ ad =bc$

Hence true.

$\therefore$ R is symmetric.

3. R to be transitive, we need to prove:

$if ((a,b),(c,d)),((c,d),(e,f)) \in R \Rightarrow ((a,b),(e,f)) \in R$

$((a,b),(c,d)) \in R$ means $ad = cb$ .... (1)

$((c,d), (e,f)) \in R$ means $fc = ed$ ...... (2)

To prove:

To be $((a,b), (e,f)) \in R$ we need to prove: $fa = be$

Multiply (1) with (2):

$adcf = bcde\\\Rightarrow fa = be$

So, R is transitive as well.

Hence proved that R is an equivalence relation.

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What is the value of b(10) when b(x)=|3-x|+1 ?

vredina [299]

B(x) = |3-x|+1

b(10) means that we need to replace x by 10:
b(10) = |3-10|+1
As you can notice, 3-10 is between the lines of the absolute power. So if the answer is positive, you keep positive. If the answer is negative, you transform it to positive. |a| = a and |-a| = a. So |a|=|-a|.
So |3-10| = |-7| = |7| = 7.

b(10) = |3-10|+1 = 7 + 1 = 8.

So the value of b(10) when b(x)=|3-x|+1 is 8.

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