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anastassius [24]
3 years ago
11

Α²+α²=c² solve for c

Mathematics
2 answers:
Irina18 [472]3 years ago
5 0

Answer:

unsolvable

Step-by-step explanation:

because we have no data on what the decimals are worth. If I'm wrong please prove me wrong, but I'm pretty sure I'm right.

bixtya [17]3 years ago
4 0
You can’t solve for c^2 without a^2
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The graph of the line x=-4 is a horizontal line
natita [175]

Answer:

Yes

Step-by-step explanation:

5 0
2 years ago
3x-8y=24 <br> y=x-8<br> Please show <br> work
In-s [12.5K]

Answer:

3x - 8y = 24   --------(1)

y = x - 8  ---------(2)

substitute  y in (1)

3x - 8( x - 8) = 24

3x - 8x + 64 = 24

-5x = 24 - 64

-5x = -40

5x = 40

x= 8

substitute x in (2)

y = 2 - 8

y = - 6

4 0
2 years ago
Read 2 more answers
g If a hypothesis test with a significance level of α rejects H0: μ1=μ2 in favor of Ha: μ1≠μ2, then thecorresponding (1- α)% con
Effectus [21]

Answer:

The statement, (1- <em>α</em>)% confidence interval for (μ₁ - μ₂) does not contain zero is TRUE.

Step-by-step explanation:

The hypothesis for a test is defined as follows:

<em>H</em>₀: μ₁ = μ₂ vs. <em>H</em>ₐ: μ₁ ≠ μ₂

It is provided that the test was rejected st the significance level <em>α</em>%.

If a decision is to made using the confidence interval the conditions are:

If the null hypothesis value is not included in the (1 - <em>α</em>)% confidence interval then the null hypothesis will be rejected and vice versa.

In this case the null hypothesis value is:

<em>H</em>₀: μ₁ - μ₂ = 0.

If the value 0 is not included in the (1 - <em>α</em>)% confidence interval for the difference between two means, then the null hypothesis will be rejected.

Thus the statement, (1- <em>α</em>)% confidence interval for (μ1- μ2) does not contain zero is TRUE.

5 0
3 years ago
Let $f(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equ
Mademuasel [1]
We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.

Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1

As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
It turns out that:
F(2) = 3/4
F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40

Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
 we see that :
x → 1 - 1/x → 1/(1-x) →x

so that:
g(g(g(x))) = x.

Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)

As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x

Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.

7 0
2 years ago
PLEASE HELP ME<br> Which expression is equivalent to –8(–5k+4)+6k
zheka24 [161]

Answer:

the last option (-5k+4)-8 + 6k

Step-by-step explanation:

-8(-5k+4)+6k

to simplify, begin by distributing the -8 so that parentheses are removed:

=(-8)(-5k) + (-8)(4) + 6k

=40k - 32 + 6k

combine 'like terms':

46k-32

the only answer which simplifies to 46k-32 is the last option:

the only difference between the original problem and the last option is that the -8 value is behind the parentheses instead of in front of it; it simplifies to be 46k-32

6 0
2 years ago
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