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3 years ago

Can anyone help me with this?

Mathematics

1 answer:

Iteru [2.4K]3 years ago

3 0

Look at the attached picture ⤴

Hope it will help u ..

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What is othorgraphic

valentinak56 [21]

The conventional spelling system of a language.

3 0

3 years ago

PLEASE OF YOU CAN SHOW THE WORK:

Alex777 [14]

Answer:

$x\approx24^\circ$

Step-by-step explanation:

We know that r and s have a combined length of 70. Therefore:

$r+s=70$

Notice that we can determine r by using the Pythagorean Theorem. In this case, r is the hypotenuse, and 20 and 8 are the legs. Therefore:

$a^2+b^2=c^2$

Substituting 20 and 8 for a and b and r for c yields:

$20^2+8^2=r^2$

Compute:

$464=r^2$

Therefore:

$r=\sqrt{464}=\sqrt{16\cdot 29}=4\sqrt{29}$

Now, we can determine s. We know that:

$s+r=70$

So, by substitution:

$s+4\sqrt{29}=70$

Therefore:

$s=70-4\sqrt{29}$

Now, notice that, with respect to x, 20 is the opposite side and s is the hypotenuse.

Therefore, we can use the sine ratio. The sine ratio is the ratio between a right triangles opposite side to its hypotenuse.

In this case, the opposite to x is 20, and the hypotenuse is s, or 70-4√29. Therefore:

$\displaystyle \sin(x^\circ)=\frac{20}{s}$

By substitution:

$\displaystyle \sin(x^\circ)=\frac{20}{70-4\sqrt{29}}$

Take the inverse sine of both sides:

$\displaystyle x^\circ=\sin^{-1}\Big(\frac{20}{70-4 \sqrt{29}} \Big)$

Use a calculator. Therefore:

$x\approx24.37^\circ\approx24^\circ$

8 0

3 years ago

Two side of a triangle have the following measure 11 and 12. Find the range of possible measure for the third side.

HACTEHA [7]

3 i’m pretty sure !

4 0

4 years ago

What 57 rounded to the nearest 100

Natali5045456 [20]

57 would be rounded to the number 100

4 0

3 years ago

Read 2 more answers

Plz help me with this

stiks02 [169]

Answer: FALSE

Step-by-step explanation:

$\text{Pythagorean formula is } (x)^2+(y)^2=(hypotenuse)^2\\\text{Pythagorean triple states that each leg and the hypotenuse are integers.}\\\\\text{Given:}\\leg_1=x^2-y^2 \qquad \qquad leg_2=2xy\qquad \qquad hypotenuse=x^2+y^2\\\\\text{Let x = 16 and y = 2k + 1 (the odd leg).}\\\text{Need to show that the hypotenuse is an integer:}$

$(16)^2+(2k+1)^2=(hypotenuse)^2\\256+4k^2+4k+1=(hypotenuse)^2\\4k^2+4k+257=(hypotenuse)^2\\4(k^2+k+64.25)=(hypotenuse)^2\\k^2+k+64.25\text{ is not a perfect square because }\sqrt{64.25} \text{ is irrational}\\\text{Therefore, the hypotenuse is not an integer.}\\\\\text{The statement is FALSE.}$

3 0

3 years ago