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Kobotan [32]
3 years ago
6

I really cannot figure out how to find x.

Mathematics
2 answers:
Musya8 [376]3 years ago
8 0

Answer:

<h3><u>4</u></h3>

Step-by-step explanation:

We know that,

cos  \alpha  =  \frac{side \: adjacent \: to \:  \alpha }{hypotenuse}  \\

<h3>Therefore, from the figure:-</h3>

cos60° = BC/ AC

=> 1/2 = 2/x

=> x = 4

katovenus [111]3 years ago
3 0
I think the answer is x=4
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The coordinate −6 has a weight of 3, and the<br> coordinate 2 has a weight of 1
Damm [24]

The weight average of the coordinates is -4

<h3>How to determine the weight average?</h3>

The complete question is given as:

The coordinate -6 has a weight of 3 and the coordinate 2 has a weight of 1. And we need to calculate the weight average

The given parameters are:

  • Coordinate -6 has a weight of 3
  • Coordinate 2 has a weight of 1.

The weight average is then calculated as:

Weight average = Sum of (Weigh * Coordinate)/Sum of Weights

So, we have:

Weight average = (-6 * 3 + 2 * 1)/(3 +1)

Evaluate the products

Weight average = (-18 + 2)/(3 +1)

Evaluate the sum

Weight average = -16/4

Evaluate the quotient

Weight average = -4

Hence, the weight average of the coordinates is -4

Read more about average at

brainly.com/question/20118982

#SPJ1

<u>Complete question</u>

The coordinate -6 has a weight of 3 and the coordinate 2 has a weight of 1. Calculate the weight average

4 0
2 years ago
1) Slope =-1, y-intercept = 0
Dima020 [189]

Answer:

1. C       2.D       3.D        4.A

Step-by-step explanation:

The first one would be C because the slope is -1, which can also be -x and the y-intercept is 0, meaning you don't need to add or subtract anything.

The second one is D because the slope is -3, meaning it's -3x, and the y-intercept is positive, meaning you add 5.

The third one is D because the slope is 5/4, which is 5/4x and the y-intercept is positive 3, meaning you add 3.

And the fourth one is A because there is no slope, so y would just equal 3.

6 0
3 years ago
3x + 2(4x - 4) = 3<br> What’s the answer?
Darya [45]
3x+2(4x-4)=3 - remove parentheses
3x+ 8x-8=3 - collect like terms
11x- 8 =3 - move the constant to the right
11x=3+8 - calculate
11x = 11 - divide both sides by 11
x= 1
3 0
3 years ago
NEED ANSWER FAST. 8th grade math
Kipish [7]
Answer: C

I can’t explain it very well, but the answer is C
6 0
3 years ago
Read 2 more answers
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
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