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3 years ago

Help will give branliest 15 pointss..

Mathematics

2 answers:

Ber [7]3 years ago

7 0

Answer:

x = 3

Step-by-step explanation:

2x + 3 = 9

2(3) + 3 = 9

6 + 3 = 9

Hope this helps!

svp [43]3 years ago

7 0

Answer: x=4

Step-by-step explanation: do the problem backward. 9-3=6, 6 divided by 2 is 4.

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Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily,

kipiarov [429]

9514 1404 393

Answer:

annually: 9.01 years
monthly: 8.69 years
daily: 8.67 years
continuously: 8.66 years

Step-by-step explanation:

For interest compounded in discrete intervals, the formula is ...

A = P(1 +r/n)^(nt)

We want to find t for P=1 and A=2, so we have ...

2 = (1 +r/n)^(nt)

ln(2) = nt·ln(1+r/n)

t = ln(2)/(n·ln(1+r/n))

A table of values for r=0.08 is attached.

For continuous compounding, the formula is ...

A = Pe^(rt)

t = ln(A/P)/r = ln(2)/0.08 ≈ 8.66434 . . . . years

annually: 9.01 years
monthly: 8.69 years
daily: 8.67 years
continuously: 8.66 years

5 0

2 years ago

Which statement describes why AABC and ADEF are congruent by SAS?

Amanda [17]

Answer:It is The first one.

Step-by-step explanation:

8 0

2 years ago

Given that x = 7.4 m and 0 = 26°, work out AB rounded to 3 SF.

Vladimir [108]

Answer:

3.24m

Step-by-step explanation:

SOH

sin(26)=O/7.4

O=7.4 sin(26)

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7 0

3 years ago

Dr. Green has to multiply the weight of the team’s bug spray bottles, 7 x 90 grams. To solve this, he writes out the equation 7

Dominik [7]

It works because 7 x 9 x 10 is the same thing as 7 x 90, since 9 x 10 is just 90. So basically he is just writing more but the answer will still be the same.

4 0

3 years ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$

This is what the Bilateral Theorem says:

$\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L$

4 0

4 years ago