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Cloud [144]
3 years ago
13

Please help me I’m doing a test and it’s very hard to me please it’s worth 200 point of my grade

Mathematics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

y = 15x

Step-by-step explanation:

For every hour that has gone by, Marco biked 15 miles, therefore, the equation of the line that shows the relationship between the time, x, and the distance, y, is <em>y = 15x</em>.

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(-w-6)/(w^2+4w-12), how do i write this into simplest form?
PIT_PIT [208]
First factorise the top = -(w+6)
The first factorise the bottom: (w^2 + 4w -12) = (w+6) (w-2)
Therefore the two (w+6) cancels out and you are left with -1 / w-2
7 0
3 years ago
What’s 7/12+1/3 in fractions show your work
uysha [10]
The answer is 11/12. (:
5 0
3 years ago
Read 2 more answers
What is the quotient of 1 1/3 and 5/6
kramer
It would be 3.667 for 11/3 and for 5/6 it would be 0.8333333333
5 0
3 years ago
If m∠CED = 56° and m∠AEB = (7x – 8)° then what is m∠BEC?
fredd [130]

Using the definition of angles on a straight line, the measure of angle BEC is: 62°.

<h3>What are Angles on a Straight Line?</h3>

Angles that lie on a straight line add up to 180 degrees.

Thus, given that EB bisects ∠AEC, therefore:

m∠BEC = 1/2(180 - m∠CED)

Plug in the values

m∠BEC = 1/2(180 - 56)

m∠BEC = 1/2(124)

m∠BEC = 62°

Thus, using the definition of angles on a straight line, the measure of angle BEC is: 62°.

Learn more about angles on a straight line on:

brainly.com/question/24024505

8 0
3 years ago
how much antifreeze 25% antifreeze and 50% antifreeze should be combined to give 40 gallon of 30% antifreeze?​
VARVARA [1.3K]

32 gallons of 25 % antifreeze should be combined with 8 gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze

<h3><u>Solution:</u></h3>

From given,

Final mixture is 40 gallon

Let "x" be the gallon of 25 % antifreeze

Then, (40 - x) is the gallon of 50 % antifreeze

Therefore, according to question,

"x" gallons of 25 % antifreeze should be combined with (40 - x) gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze

<h3><u>Therefore, we frame a equation as:</u></h3>

25 % of x + 50 % of (40 - x) = 30 % of 40

Solve for "x"

\frac{25}{100} \times x + \frac{50}{100} \times (40-x) = \frac{30}{100} \times 40\\\\0.25x + 0.5(40-x) = 12\\\\0.25x + 20 - 0.5x = 12\\\\0.25x - 0.5x = 12 - 20\\\\-0.25x = -8\\\\0.25x = 8\\\\Divide\ both\ sides\ by\ 0.25\\\\x = 32

Thus, 32 gallons of 25 % antifreeze is used

Then, (40 - x) = (40 - 32) = 8

Thus 8 gallons of 50 % antifreeze is used

6 0
3 years ago
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