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Scorpion4ik [409]

3 years ago

11

You can complete three-tenths of a task in an hour. Your friend can complete one-fifth of the same task in an hour. How long wil

l it take to complete the task if you work together?

1 answer:

iren [92.7K]3 years ago

4 0

Answer:

together complete the task in 2 hours

Step-by-step explanation:

given, you can complete three-tenths of task in one hour

you can complete a task in 3.33 hour

your friend can complete one-fifth of task in one hour

your friend can complete a task in 5 hour

together complete the task in

= $\frac{3.33\times 5}{3.33+5}$

= 2 hour answer

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Which shows one way to determine the factors of x3 – 9x2 + 5x – 45 by grouping?

Valentin [98]

Factoring by grouping usually pairs up the first 2 sets of expressions with the second 2 sets. Ours looks like this, then: $(x^3-9x^2)+(5x-45)=0$ . If we factor out the common x-squared in the first set of parenthesis, along with factoring out the common 5 in the second set, we get this: $x^2(x-9)+5(x-9)$ . Now the common expression that can be factored out is the (x-9). When we do that, here's what it looks like: $(x-9)(x^2+5)$ . I'm not sure how far you are going with this. You could set each of those equal to 0 and solve for x in each case. The first one is easy. If x - 9 = 0, then x = 9. The second one involves the imaginary i since x^2 = -5. In that case, $x=i \sqrt{5},-i \sqrt{5}$ . Hopefully, in what I have given you, you can find what you're looking for.

7 0

3 years ago

Prove that there is only one circle determined by three noncollinear points

tensa zangetsu [6.8K]

Answer:

O lies on the perpendicular bisector of AB. ... Then, O' will lie on the perpendicular bisector PQ and RS. We know that, two lines cannot intersect at more than one point, So O' must coincide with O. Hence, there is one and only one circle passing through three non collinear points

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

3 years ago

Graph y= 3/4x - 1/2

Elenna [48]

I hope this helps! :)

4 0

3 years ago

Read 2 more answers

How to graph an exponential function

irina1246 [14]

Step-by-step explanation:

Exponential function is given by general form $y=ab^x$

Where a and b are constants.

say a=1 and b=2 then we can write function as :

$y=1*2^x$

or $y=2^x$

To graph this or any exponential function, we just need to find some points then join them by a curved line.

Like plug x=0, 1, 2,... into above function and find points:

plug x=1

$y=2^x=2^1=2$

Hence point is (1,2)

Find more points similarly then graph them to get the graph as shown below:

7 0

4 years ago