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3 years ago

2. Is 4x – 3 = 19 true, false, or open? A)true B)false C)open

Mathematics

1 answer:

ElenaW [278]3 years ago

6 0

Answer: B. False

Step-by-step explanation: Because you do

4x - 3 = 19

+3 +3 =4x = 22

You divide 4x divide by 4 and you get x

Then you divide 22 by 4 and get 5.5

So the answer is X = 5.5

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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0

3 years ago

Which equation is equivalent to 4x+3 – 647 I don’t know how to do it

german

Answer:3.54

Step-by-step explanation:

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3 years ago

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a worker in a silver mine descends 71 feet. Use an integer to represent the change in the workers position.

alexira [117]

Answer:

iuyhgt

Step-by-step explanation:

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3 years ago

Using a directrix of y = −2 and a focus of (1, 6), what quadratic function is created? Question 3 options:

deff fn [24]

If your directrix is a "y=" line, that means that the parabola opens either upwards or downwards (as opposed to the left or the right). Because it is in the character of a parabola to "hug" the focus, our parabola opens upwards. The vertex of a parabola sits exactly halfway between the directrix and the focus. Since our directrix is at y = -2 and the focus is at (1, 6) AND the parabola opens upward, the vertex is going to sit on the main transversal, which is also the "line" the focus sits on. The focus is on the line x = 1, so the vertex will also have that x coordinate. Halfway between the y points of the directrix and the focus, -2 and 6, respectively, is the y value of 2. So the vertex sits at (1, 2). The formula for this type of parabola is $(x-h)^2=4p(y-k)$ where h and k are the coordinates of the vertex and p is the DISTANCE that the focus is from the vertex. Our focus is 4 units from the vertex, so p = 4. Filling in our h, k, and p: $(x-1)^2=4(4)(y-2)$ . Simplifying a bit gives us $(x-1)^2=16(y-2)$ . We can begin to isolate the y by dividing both sides by 16 to get $\frac{1}{16}(x-1)^2=y-2$ . Then we can add 2 to both sides to get the final equation $f(x)=\frac{1}{16}(x-1)^2+2$ , choice 4 from above.

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3 years ago

Solve 3v -12 = 15 v = -9 v = 9 v = 1 v = -1

svetoff [14.1K]

Answer:

v=9

Step-by-step explanation:

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3 years ago