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4 years ago

About how long is the log that goes across the creeks?

Mathematics

1 answer:

soldi70 [24.7K]4 years ago

6 0

Answer:

6 feet

Step-by-step explanation:

we know that

The triangles of the figure are similar by AA Similarity Theorem

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

Applying proportion

Let

x ----> the length of the log in feet

$\frac{x}{8}=\frac{9}{12}\\\\x=8(9)/12\\\\x=6\ ft$

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MAXImum [283]

Answer:

Angle A is 101

Angle B is 79

Angle C is 83

Angle D is 97

Step-by-step explanation:

Vertical angles are congruent, which applies to C.

Lines are parallel so 79 and angle B are the same

Angle a is 101 because its a supplementary angle

Angle D is 97 because its also a supplementary angle so you subtract 83 from 180.

3 0

3 years ago

Which graph correctly represents the equation y=−4x−3?

pentagon [3]

Idk if u got the answer or not, but It's the last graph. Just by looking at it you can point two things out.

1. The y intercept is -3 (so the graph needs to have the point (0,-3) that takes out two options.)

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7 0

4 years ago

Mr. Darcy could knit 4 ugly sweaters in 100 minutes. How long will it take him to knit 6 ugly holiday sweaters

Nostrana [21]

If you divide 100 by 4 which equals 25 that would be 25mins for one ugly sweater
If you times by 2 that equals 50
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3 0

3 years ago

Read 2 more answers

Using a commen denominator to order fractions, please help...

MrRa [10]

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4 years ago

We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure

STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be $h(t) = -16\cdot t^{2} + 128\cdot t + 320$ , the first and second derivatives are, respectively:

First Derivative

$h'(t) = -32\cdot t +128$

Second Derivative

$h''(t) = -32$

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

$-32\cdot t +128 = 0$

$t = \frac{128}{32}\,s$

$t = 4\,s$ (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

$h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320$

$h(4\,s) = 576\,ft$

The highest altitude that the object reaches is 576 feet.

6 0

4 years ago