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Leno4ka [110]
3 years ago
12

Will give brainlst Have a nice day

Mathematics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

Remember that the Pythagorean's theorem says that:

For a triangle rectangle with hypotenuse H and catheti A and B:

H^2 = A^2 + B^2

Here we also need to remember that the area of a square of side length L is:

area = L^2

Now let's solve this.

First, we start with two squares, one of side length a and the other of side length b.

Such that the complete area in the first image is:

area = a^2 + b^2

Now we draw two triangle rectangles with catheti a and b, and with hypotenuse c.

in step 3, we rotate those triangles in order to make a larger square, with side length c, with an area equal to:

area = c^2

Notice that we never added more shapes, so the area of the image did not change in all this process, then the initial area must be equal to the final area:

a^2 + b^2 = area = c^2

a^2 + b^2 = c^2

And remember that a and b are the catheti of the triangles, and c is the hypotenuse, then this is the Pythagorean's theorem.

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Zigmanuir [339]

Answer:

14

Step-by-step explanation:

6 (1) + 4(-3) + -1 = -7

A = 1

B = -3

C = -1

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Plugged it in:

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2 + -12 + 24 = 14

5 0
4 years ago
Evaluate bcl, given a = 5, b=-3, and c= -2.
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Answer:

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5 0
3 years ago
Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The m
svp [43]

Answer:

As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.

Step-by-step explanation:

As we suspect the variability of concentration F - test is applied.

n1=10    s1=4.7

n2=16      s2=5.8

And α = 0.05.

The null and alternate hypothesis are

H0: σ₁²=σ₂²       Ha:  σ₁²≠σ₂²

The null hypothesis is  the variability in concentration does not  differ for the two companies.

against the claim

the variability in concentration may differ for the two companies

The critical region F∝(υ1,υ2) = F(0.025)9,15= 3.12

and 1/F∝(υ1,υ2) = 1/3.77= 0.26533

where υ1= n1-1= 10-1= 9 and υ2= n2-1= 16-1= 15

Test Statistic

F = s₁²/s₂²

F= 4.7²/5.8²=0.6566

Conclusion :

As the calculated F lies in the acceptance region therefore we conclude that there is not sufficient evidence to support the claim that the variability in concentration may differ for the two companies. Hence Ha is rejected and H0 is accepted.

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3 years ago
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The distributive property combines like terms.

6 0
4 years ago
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think of it like this, after 1 hour there is 2 inches of rain, after 2 hours there is 4 inches of rain, and after 10, there should be 20, so, as it runs 1 it should rise 2.
5 0
3 years ago
Read 2 more answers
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