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3 years ago

Please Help Me To Solve For X

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

8 0

Answer:

u & v = 8

Step-by-step explanation:

This triangle is a special 45-45-90 triangle!

In a 45-45-90 triangle, to solve for the hypotenuse use the formula

$h=(sidelength)*\sqrt{2}$

Here we see that 8 is being multiplied by radical 2!

We can prove that 8 is the side length by using the Pythagorean Theorem!

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- 7th Grade Work - Please write your answer like this: 1. 2. 3.

Naily [24]

Answer:

1. -4x < 8

2. 4x < 8

3. When something is done to one side of the equation, the same thing must be done to the other side. Dividing each side by -4 in part A would cause the sign to flip because the sign is always flipped when dividing by a negative. Because 4 in 4x (part B) is not negative, the sign is not flipped.

Step-by-step explanation:

Hope this helped!!

8 0

3 years ago

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Here have some points

kozerog [31]

Answer:

Cheers bro ;)

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3 years ago

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

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3 years ago

10.4 sweets cost 28p altogether, how much much do 7 sweets cost

Serjik [45]

28/10.4=2.69
2.69*7=18.83

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3 years ago

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A ballroom in a hotel has the shape shown in the diagram. The hotel manager needs to determine how much carpet is needed to reca

Aleonysh [2.5K]

We can separate the shape into a rectangle and a square
the area of the rectangle :
$3 \times 12 = 36$
the area of the square:
${6}^{2} = 36$
the area of the whole shape:
$36 + 36 = 72$
hope this helps

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3 years ago

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