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2 years ago

What is the value of x in equation 0.2(x+1)+ 0.5x = - 0.3(x-4)

Mathematics

2 answers:

katen-ka-za [31]2 years ago

8 0

Answer:

-3.5

Step-by-step explanation:

i just worked it out

Greeley [361]2 years ago

6 0

Answer: x=1

Step-by-step explanation:

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Highest comman factor of 200 and 240

Schach [20]

Answer:

Step-by-step explanation:

GCF of 200 and 240 is 40

4 0

2 years ago

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What is (13x + 9k) + (17x + 6k), simplified?

Mazyrski [523]

The answer is 15x+30k

7 0

2 years ago

I need some help on this math problem, thanks so much

bekas [8.4K]

Answer:

The answer is C.

Step-by-step explanation:

Hit 'em with the Law of Sines.

sin(A)/a = sin(B)/b.

Let's say x is equal to "A", thus 5 is "a".

sin(x)/5 = sin(B)/b.

We could go for the obvious choice for "B", which would be the 90 degrees shown. To solve for the hypotenuse which will be "b", let's use the Pythagorean Theorem:

a^2 + b^2 = c^2

5^2 + 20^2 = c^2

25 + 400 = 425

sqrt(425) = about 20.6, which we can now substitute "b" with.

sin(x)/5 = sin(90)/20.6

sin(x)/5 = 1/20.6

sin(x)/5 = 0.04854...

sin(x) = 0.2427...

You can plug in sin^-1(0.2427) into your calculator, and you should end up with something like 14.047... which equates to answer choice C.

6 0

2 years ago

PLEASE HELP ASAP DUE IN 5 MIN PLEASEEEE !!!!!! Which best compares the maximum value of the two functions?

professor190 [17]

Answer:

3 is the correct answer

Step-by-step explanation:

because

5 0

1 year ago

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

2 years ago