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3 years ago

If x and y are 2 complementary angles then find the value of y if measure of angle x is 20 degree

Mathematics

1 answer:

schepotkina [342]3 years ago

6 0

Answer:

y=70 degrees

Step-by-step explanation:

If 2 angles are complimentary, they add up to 90 degrees.

x+y=90 degrees

x=20 degrees

y=90-20 degrees

y=70 degrees

If that is wrong, try this. Let's say the questions gives us the fact that x is 20 degrees greater than y.

y=x

Substituting x for y, the equation is x+x+20=90

Since y is x, x would also be y, which is 35. then x+20 would be 55.

So y would be 35.

THE ANSWER WOULD BE 70 DEGREES, BUT IF THAT IS WRONG, DO THE SECOND ONE, WHICH IS 35

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Solve for x. Show your work with proper statements and notation. a Point O ∈ AB

viktelen [127]

Answer:

x=27

Step-by-step explanation:

m<AOB = 180

36+90+2x=180

x=27 degrees

4 0

3 years ago

What is the value of "c" in the quadratic equation 3x 2 + 5x + 7 = 0? 3 5 7

dmitriy555 [2]

Answer:

$c=7$

Step-by-step explanation:

They are just asking you to compare

$ax^2+bx+c=0$ to

$3x^2+5x+7=0$ .

What constant values are in the place of $a,b, \text{ and } c$ .

$a=3$

$b=5$

$c=7$

7 0

4 years ago

Read 2 more answers

The rod is made of A-36 steel and has a diameter of 0.22 in . If the rod is 4 ft long when the springs are compressed 0.7 in . a

ziro4ka [17]

The force in the rod when the temperature is 150 °F is 718.72 pounds-force.

<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>

Let suppose that rod experiments a quasi-static deformation and that both springs have a linear behavior, that is, force ( $F$ ), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to temperature increase creates a spring deformation additional to that associated with mechanical contact.

Given simmetry considerations, we derive an expression for the spring force ( $F$ ), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:

$F = k\cdot (\Delta x + 0.5\cdot \Delta l)$ (1)

Where:

$k$ - Spring constant, in pounds-force per inch.
$\Delta x$ - Spring deformation, in inches.
$\Delta l$ - Rod elongation, in inches.

The rod elongation is described by the following thermal dilatation formula:

$\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})$ (2)

Where:

$\alpha$ - Coefficient of linear expansion, in $\frac{1}{^{\circ}F}$ .
$L_{o}$ - Initial length of the rod, in inches.
$T_{o}$ - Initial temperature, in degrees Fahrenheit.
$T_{f}$ - Final temperature, in degrees Fahrenheit.

If we know $k = 1000\,\frac{lb}{in}$ , $\Delta x = 0.7\,in$ , $\alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}$ , $L_{o} = 48\,in$ , $T_{o} = 30\,^{\circ}F$ and $T_{f} = 150\,^{\circ}F$ , then the force in the rod at final temperature is:

$F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]$

$F = 718.72\,lbf$

The force in the rod when the temperature is 150 °F is 718.72 pounds-force. $\blacksquare$

To learn more on deformations, we kindly invite to check this verified question: brainly.com/question/13774755

3 0

3 years ago

PLEASE HELP ME!! DUE IN 10 MINS!!!

GuDViN [60]

The answer is 553 general admission tickets sold and 295 VIP tickets sold
Since this is due in 10 minutes, I didn’t give a step by step explaination, but if you need one, just comment on this.

6 0

3 years ago

Consider the following system of equations:

Umnica [9.8K]

X=3 and the work to get the answer is attached.

4 0

3 years ago