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2 years ago

Find XS, VY, and YZ. I am very confused on what to do.

Mathematics

1 answer:

Dafna1 [17]2 years ago

7 0

Answer:

Do you see the lines and how they are named? I am going to try and explain this in less words than I can. Where it says XS you have to match it up with the lines. But if it says that you have to find XSU (which it does not) you would have to go up all the way but you stop where it says the S so you wont go all the way up to the U. I hope this helps!

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Compare: 10 quarts and 2.5 gallons *

Charra [1.4K]

10 quarts = 2.5 gallons

8 0

2 years ago

What is the Value Of X,

Zina [86]

The value of x in the figure is 9.72

<h3>How to determine the value of x?</h3>

The given shape is a triangle, and the value of x can be calculated using the following laws of cosine

a^2 = b^2 + c^2 - 2bc * cos(a)

So, the equation becomes

x^2 = 14^2 + 10^2 - 2 * 14 * 10 * cos(44)

Evaluate the value of cos(44)

x^2 = 14^2 + 10^2 - 2 * 14 * 10 * 0.7193

Evaluate the product

x^2 = 296 - 201.404

Evaluate the difference

x^2 = 94.596

Evaluate the exponent

x = 9.72

Hence, the value of x in the figure is 9.72

Read more about laws of cosine at

brainly.com/question/4372174

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6 months ago

Where are the asymptotes for the following function located? f (x) = StartFraction 14 Over (x minus 5) (x + 1) EndFraction

m_a_m_a [10]

Answer:

2 vertical asymptotes occurring at x = 5 and x = -1

Step-by-step explanation:

given

$f(x) = \frac{14}{(x-5)(x+1)}$

recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)

Equating the denominator to zero,

(x-5)(x+1) = 0

(x-5) =0

x = 5 (first asymptote)

or (x+1) = 0

x = -1 (2nd asymptote)

6 0

2 years ago

Solve this for Brainliest!<br><br>See the attachment [Level - GCSE]

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

<u>Using Pythagoras</u>

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

<u>Using quadratic formula</u>

<u /> $x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$

$x = 3 + 4\sqrt{3} = 9.928(4s.f.)$

$x = 3 - 4\sqrt{3} = -3.928(4s.f.)$ (Length can't be negative)

∴ $x = 3 + 4\sqrt{3}$

3 0

1 year ago

Read 2 more answers

At what point of the curve y = cosh(x does the tangent have slope 2?

Dahasolnce [82]

The hyperbolic cos (cosh) is given by
cosh (x) = (e^x + e^-x) / 2
The slope of a tangent line to a function at a point is given by the derivative of that function at that point.
d/dx [cosh(x)] = d/dx[(e^x + e^-x) / 2] = (e^x - e^-x) / 2 = sinh(x)
Given that the slope is 2, thus
sinh(x) = 2
x = sinh^-1 (2) = 1.444

Therefore, the curve of y = cosh(x) has a slope of 2 at point x = 1.44

5 0

2 years ago

Find XS, VY, and YZ. I am very confused on what to do.​

Find XS, VY, and YZ. I am very confused on what to do.