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Arada [10]
3 years ago
12

What answer do you get when you solve 0.4(2a-0.3b)

Mathematics
1 answer:
guapka [62]3 years ago
5 0

Answer:

0.8a-0.12b

Step-by-step explanation:

Use the distributive property:

0.8a-0.12b

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Use graph to answer question
slamgirl [31]

Answer:

cos(θ) = 3/5

Step-by-step explanation:

We can think of this situation as a triangle rectangle (you can see it in the image below).

Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.

Here we want to find cos(θ).

You should remember:

cos(θ) = (adjacent cathetus)/(hypotenuse)

We already know that the adjacent cathetus is equal to 3.

And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:

3^2 + 4^2 = H^2

We can solve this for H, to get:

H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5

The hypotenuse is 5 units long.

Then we have:

cos(θ) = (adjacent cathetus)/(hypotenuse)

cos(θ) = 3/5

5 0
3 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
1 year ago
Enter the correct answer pls help​
Phoenix [80]

Answer:

\huge\boxed{\sf x = 8a}

Step-by-step explanation:

\sf \frac{3}{a} x - 4 = 20\\\\Add \ 4 \ to \ both \ sides\\\\\frac{3}{a} x = 20+4\\\\\frac{3}{a} x = 24\\\\Multiply \ both \ sides \ by \ a\\\\3x = 24 * a\\\\Divide \ 3 \ to \ both \ sides\\\\x = 24a / 3\\\\x = 8a\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
4 0
3 years ago
<img src="https://tex.z-dn.net/?f=95%20%5Cdiv%2013.246" id="TexFormula1" title="95 \div 13.246" alt="95 \div 13.246" align="absm
Alexxx [7]

Answer:

  • 7.1719764457194
  • is the answer
7 0
2 years ago
Read 2 more answers
I need help with this please!!
topjm [15]

Answer:

8(3j -2)

4(6j -4)

Step-by-step explanation:

24j - 16

Rewriting

8*3*j - 8*2

Factor out the greatest common factor

8(3j -2)

We can also factor out 4

4*6j - 4*4

4(6j -4)

3 0
2 years ago
Read 2 more answers
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