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3 years ago

1 - For what value of n is 20 = n/5 + 10 ?

1 answer:

Elena-2011 [213]3 years ago

5 0

First one

20=n/5+10
minus 10 both sides
10=n/5
times 5 both sides
50=n

2.
hmm
expand
p(x)=2x^2+5x+3-3x-3n
p(x)=2x^2+2x+3-3n
it looks like the coefients of the x terms are divisble by 2
because
if we have
p(x)=2(something)
the p(x) is divisible by 2

therefor that factor is 2 or -2
find what value of n works
so
3-3n=a multipule of 2
we are given
-3,-2,0,2
try each
3-3(-3)=3+9=12, that's divisble by 2
3-3(-2)=3+6=9, that's not divisible by 2
3-3(0)=3, that's not divisble by 2
3-3(2)=3-6=3, that's not divisble by 2

the onnly one that woks is when n=-3

answer is A

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3 years ago

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ivanzaharov [21]

Answer:

I) Eq(1) reason: sum of segments of a straight line

II) Eq(2) reason: Given PQ = ST & PS = RT

III) Eq(3) reason: sum of segments of a straight line

IV) Eq(4) reason: Same value on right hand sides of eq(2) and eq(3) demands that we must equate their respective left hand sides

V) Eq(5) reason: Usage of collection of like terms and subtraction provided this equation.

Step-by-step explanation:

We are given that;

PS = RT and that PQ = ST

Now, we want to prove that QS = RS.

From the diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

PQ + QS = PS - - - - (eq 1)

Now, from earlier we saw that PQ = ST & PS = RT

Thus putting ST for PQ & PS for RT in eq 1,we have;

ST + QS = RT - - - - (eq 2)

Again, from the line diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

RS + ST = RT - - - - -(eq 3)

From eq(2) & eq(3) we can see that both left hand sides is equal to RT.

Thus, we can equate both left hand sides with each other to give;

ST + QS = RS + ST - - - (eq 4)

Subtracting ST from both sides gives;

ST - ST + QS = RS + ST - ST

This gives;

QS = RS - - - - (eq 5)

Thus;

QS = RS

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