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3 years ago

Round to the nearest tenth.

Mathematics

1 answer:

inessss [21]3 years ago

6 0

Answer: Angle A = 53.9 degrees

Step-by-step explanation: We have a right angled triangle with two sides clearly given and one angle to be calculated. If the angle to be calculated is angle A, then having angle A as our reference angle, line AC (10 units) is the adjacent, line CB is the opposite while line AB (17 units) is the hypotenuse. Having been given the adjacent and the hypotenuse, we can now use the trigonometric ratio as follows;

CosA = adjacent/hypotenuse

CosA = 10/17

CosA = 0.5882

By use of the calculator or table of values,

A = 53.97

Approximately to the nearest tenth,

A = 53.9 degrees

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The difference between twice a number and 2

Pie

Answer:

2 is a number and "twice a number" means multiply by 2.

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3 years ago

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Find the distance between (2, 2) and (-4, 4). Round answer to the nearest tenth.

Naya [18.7K]

Answer:

6.3

Step-by-step explanation:

First off, we need the distance formula, which is:

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

If we plug in the points, we get:

$\sqrt{(-4 - 2)^2 + (4 - 2)^2}$

If we simplify everything under the square root, we get:

$\sqrt{(-6)^2 + (2)^2}$

$\sqrt{36 + 4}$

$\sqrt{40}$

In decimal, the answer is 6.3

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3 years ago

Solve for x<br>2x²-3x-209=0

kifflom [539]

Answer:

x= 11

Step-by-step explanation:

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3 years ago

Two interior angles in a triangle measure 101 degrees and 34 degrees. What is the measure of the third angle?

Colt1911 [192]

All the angles in a triangle equal 180 degrees.

So,

101+34+x=180

x= the missing angle.

135+x=180

Subtract both sides by 135.

x= 45

The missing angle equals 45 degrees.

I hope this helps!
~kaikers

5 0

3 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers:

$\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}$

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

$undefined$

6 0

1 year ago