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3 years ago

Statement A:water contains dissolved minerals.

Mathematics

2 answers:

Doss [256]3 years ago

8 0

Answer:

Converse

Step-by-step explanation:

dimaraw [331]3 years ago

7 0

This statement is all the above

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What is the y-intercept of the quadratic function f(x)=x^2-5x+6?

tatiyna

Answer:

Step-by-step explanation:

f(x)=x^2-5x+6

To find the y intercept, set x=0

f(0) = 0^2 -5(0) +6

f(0) = 6

The y intercept =6

6 0

3 years ago

A 32-m tall building casts a shadow. The distance from the top of the building to the tip of the shadow is 35 m. Find the length

Vlad [161]

Answer:

sqr 201

Step-by-step explanation:

use the Pythagorean theorem

35 is the hypotenuse so 35^2-32^2=x^2

x= sqr 201

3 0

3 years ago

Solve for x. Give your answer in simplest radical form

Tasya [4]

Use the Pythagorean Theorem to find the length of the side marked "x."
x^2 + (10 mi)^2 = (13 mi)^2. Thus, x^2 + 100 mi^2 = 169 mi^2.
Next, x^2= (169-100) mi^2, or x^2 = 69 mi^2. Find the positive square root of both sides of this equation. What is it?

3 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$