A palm tree casts a 24 foot shadow. The end of the shadow is 44 feet from the top of the tree. How tall is the tree? Pls help

Which statements can be concluded from the diagram and used to prove that the triangles are similar by the SAS similarity theore

velikii [3]

Answer:

RS/VU=ST/UT and ∠S≅∠U

Step-by-step explanation:

we know that

The <u>Side-Angle-Side Similarity Theorem </u>states that: If two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar

In this problem the included angle is

∠S≅∠U

therefore

side RS must be proportional to side VU and side ST must be proportional to side UT

so

RS/VU=ST/UT

<em>Verify</em>

substitute the given values

12/6=16/8

2=2 -----> is true

therefore

The two sides are proportional

6 0

2 years ago

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If 75 is 25% of a value, what is that value?

sertanlavr [38]

75÷25%
25%=0.25
75÷0.25
=300
Your answer is 300

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

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Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

2 years ago

(01.01 MC)Why is 1 + (−5) equal to −4? I NEED HELP ASAPPPP

miv72 [106K]

Answer:

Hey there!

1+(-5)=1-5

1-5 = -4

Let me know if this helps :)

3 0

2 years ago

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