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4 years ago

What is the value of p?

Mathematics

1 answer:

anzhelika [568]4 years ago

3 0

The ratio of a 45-45-90 triangle is x-x-xrad2
So we can make the equation 44=xrad2
we can see that the answer is 22rad2 so the answer is b

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Expand the given power using the Binomial Theorem. (10k – m)5

agasfer [191]

Answer:

$(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

$(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}$

∵ The binomial is $(10k-m)^{5}$

∴ a = 10k , b = -m and n = 5

∴ $(10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}$

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ $(10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})$

∴ $(10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

5 0

3 years ago

7. Given the systems below, state how many solutions there are. YOU MUST GIVE A<br> REASON

Dmitry [639]

In a there will be 1 solution as 7x and 5x + 2 is the same and in b there will be 2 solutions as they both are 2 different questions hope this helped

6 0

4 years ago

50 POINTS AWARDED PLEASE ANSWER THE QUESTION RIGHT AND YOU WILL GET BRAINLESS ANSWER A THANK YOU A FRIEND REQUEST PLEASE DON'T J

zepelin [54]

The answer is 35x=700 because you would divide 700 by 35 to get the amount of months that you've spent. So x will equal the amount of months.

6 0

3 years ago

10 POINTS

Svetradugi [14.3K]

Answer:

C, F, D

Step-by-step explanation:

For each side of each triangle, think of the side as the hypotenuse of a right triangle. Then use the Pythagorean theorem to find the length of the hypotenuse, which is the side of the triangle.

Left problem:

From point C to point A, go 2 left and 7 up.

$AC = \sqrt{2^2 + 7^2} = 7.3$

From point A to point B, go 5 right and 1 down.

$AB = \sqrt{5^2 + 1^2} = 5.1$

From point B to point C, go 6 down and 3 left.

$BC = \sqrt{6^2 + 3^2} = 6.7$

Perimeter = 7.3 + 5.1 + 6.7 = 19.1

Do the same for the other triangles.

Middle problem:

From point F to point D, go 2 left and 7 up.

$FD = \sqrt{2^2 + 7^2} = 7.3$

From point D to point E, go 7 right and 2 down.

$DE = \sqrt{7^2 + 2^2} = 7.3$

From point E to point F, go 5 down and 5 left.

$EF = \sqrt{5^2 + 5^2} = 7.1$

Perimeter = 7.3 + 7.3 + 7.1 = 21.7

Right problem:

From point L to point J, go 7 left and 1 up.

$LJ = \sqrt{7^2 + 1^2} = 7.1$

From point J to point K, go 4 up and 5 right.

$JK = \sqrt{4^2 + 5^2} = 6.4$

From point K to point L, go 2 right and 5 down.

$KL = \sqrt{2^2 + 5^2} = 5.4$

Perimeter = 7.1 + 6.4 + 5.4 = 18.9

4 0

3 years ago

2X minus one equals five

stepan [7]

2x-1=5
2(3)-1=5
6-1=5
5=5

x=3

7 0

3 years ago

Read 2 more answers