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Snezhnost [94]
4 years ago
15

[x] + 10 + 1 answer algebraically

Mathematics
1 answer:
Svetlanka [38]4 years ago
3 0

Combine Like Terms:

<span>=<span><span>x+10</span>+1</span></span><span><span>
</span></span><span>=<span><span>(x)</span>+<span>(<span>10+1</span>)</span></span></span><span><span><span>
</span></span></span><span>=<span>x+<span>11</span></span></span>

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3x-8&lt;-2x+22<br><br> Show it with steps please.
Neko [114]

Answer:

x < 6

Step-by-step explanation:

1. 3x+2x = 5x

2. 5x - 8 < 22

3. 22+ 8 =30

4. 5x < 30

5. 30/5 = 6

6. x < 6

8 0
3 years ago
Read 2 more answers
Lim x--&gt;0 ((sqrt(ax+b) -2) / x) =1 show steps....
Andrew [12]
Lim x→0 (√(ax+b)-2)/x=1

You want to know the value of "a" and "b"

lim x→0  (√(ax+b)-2)/x=(√(0+b)-2)/0=(√b -2)/0;

Then if (√b -2)/0=1; the numerator must be "0"
(√b-2)=0
√b=2
(√b)²=2²
b=4

It is necessary the numerator must be "0", if the denominator is "0" and the result is equal a number.

Therefore:
lim     (√(ax+4)-2)/x=1
x⇒0

I imagine you know Taylor Series.
√(ax+4)=(4(1+ax/4))¹/²=2(1+ax/4)¹/²
Remember:
               (1/2)   
(1+x)ᵃ=<span>Σ ( a  ) x^a
</span>
In our case:
                   (1/2)             (1/2)              (1/2)
(1+ax/4)¹/²=(   0) (ax/4)⁰+(  1 ) (ax/4)¹+(   2) (ax/4)²+...
                =1 +(1/2) ax/4 + -1/8 (ax/4)²+...
                =1+ax/8-a²x²/128+...

Therefore:
lim     (√(ax+4)-2)/x=lim       [2(1+ax/8-a²x²/128+...)-2]/x=
x⇒0                          x⇒0 

lim       [(2+ax/4-a²x²/64+...)-2]/x=
x⇒0 

lim      (ax/4-a²x²/64+...)/x=
x⇒0

lim   x(a/4-a²x/64+...)/x=
x⇒0

lim    (a/4-a²x/64+...)=(a/4-0-0-0-...)=4/a
x⇒0

Because:

lim     (√(ax+4)-2)/x=1
x⇒0

Then:
4/a=1 ⇒  a=4

Answer: a=4; b=4
4 0
3 years ago
Question 6 of 10
jek_recluse [69]

Answer:

c=18

Step-by-step explanation:

5 0
3 years ago
to Make 2 Batches of nut bars jayda needs to use 4 eggs . How Many Eggs are Used in each batch of nut bars .
dimulka [17.4K]
2 eggs are used for each batch of nut bars
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3 years ago
11x10 to the power of one​
zysi [14]
The answer is 110
Hope this helps :)
8 0
2 years ago
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