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2 years ago

7

PLS HELP ME WITH THIS QUESTION, IM LITERALLY SO FRUSTRATED PLSSSS, WILL MARK BRAINEST, AND PLS EXPLAIN HOW U GOT THE ANSWER PLS

PLS PLS!!!!!!

1 answer:

Veronika [31]2 years ago

8 0

Answer:

64:1

It doubles every 10 minutes, so it multiplies by two 6 times in an hour. 1x2x2x2x2x2x2 = 64

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What is the value of variable for -4(y - 2) = 12<br> Explain please

Scrat [10]

Answer:

-4(y - 2) = 12

Standard form:

−4y − 4 = 0

Factorization:

−4(y + 1) = 0

Solutions:

y = −4

4

= −1

Step-by-step explanation:

7 0

3 years ago

The 1992 world speed record for a bicycle (human-powered vehicle) was set by Chris Huber. His time through the measured 200 m st

Reil [10]

Answer:

a) 30.726m/s and b) 5.5549s

Step-by-step explanation:

a.) What was Chris Huber’s speed in meters per second(m/s)?

Given the distance and time, the formula to obtain the speed is

$v=\frac{d}{t}$ .

Applying this to our problem we have that

$v=\frac{200m}{6.509s}= 30.726m/s$ .

So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.

b) What was Whittingham’s time through the 200 m?

In a) we stated that $v=\frac{d}{t}$ . This formula implies that

$t=\frac{d}{v}$ .

First, observer that $19\frac{km}{h} =19,000\frac{m}{h}=\frac{19,000}{3,600}m/s= 5.2777m/s$ .

Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, $v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.$

Then, applying 1) we have that

$t=\frac{200m}{36.003m/s}=5.5549s.$

So, Sam Whittingham’s time through the 200 m was 5.5549s.

5 0

3 years ago

Points A,B, and C are collinear. Points M and N are the midpoints of segments AB and AC. Prove that BC = 2MN

irinina [24]

Look at the picture.

1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y

|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|

6 0

3 years ago

Read 2 more answers

(-3*10^9)(1.27*10^-12) in scientific notation

katen-ka-za [31]

It would be = -3.81 * 10^-3

7 0

3 years ago

Which statement is valid when a circumscribed circle of an acute triangle is

Anni [7]

<h3>There are two answers: Choice C, Choice D</h3>

Explanation:

For any acute triangle, the center of the circle is always inside the triangle.

For any right triangle, the center of the circle is the midpoint of the hypotenuse. The hypotenuse is the diameter of the circle.

For any obtuse triangle, the center of the circle is outside the triangle.

----------

Let's look at the answer choices more closely

A. False. This is only true if we have a right triangle as mentioned earlier.

B. False. This describes an inscribed circle.

C. True. As stated earlier.

D. True. This is based on how the circumscribed triangle is defined. It is set up to be the circle that goes through all three vertex points of a triangle. In other words, it is the smallest circle to completely enclose the given triangle.

4 0

2 years ago