All categories

2 years ago

I DONT UNDERSTANDDD!!!!!!!!!!!!

Mathematics

1 answer:

EleoNora [17]2 years ago

6 0

You put the points on the line for example .5 will go on point .5 on the line etc

You might be interested in

Which of the following are valid names for the triangle below? Check all that

Bingel [31]

The answer is e because I look it up

7 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

3 years ago

Debbie set a goal to run 6 miles. On Monday she ran 2.6 miles and on Tuesday she ran 1.75 miles. How far would Debbie need to ru

Makovka662 [10]

She will need to run 1.65 to reach her go

7 0

3 years ago

What is everyone being for halloween.

Vikentia [17]

I'm going to be a A Killer Nurse. How about you?

6 0

3 years ago

Read 2 more answers

Help with this question.

pychu [463]

Answer:

$angleS=83^o$

$angleR=23^o$

Step-by-step explanation:

<h3><u>The Law of Sines</u></h3>

The Law of Sines states that for a given triangle (in a plane), the ratio formed by the Sine of any one of the three interior angles and the side across from it is equal to a common number.

If a triangle is drawn in standard form, with the three vertices identified as A, B, and C, and the interior angles at each vertex simply identified as angleA, angleB, angleC, the sides across from those angles are identified as a, b, and c, respectively. Given such a labeling for a triangle, the Law of Sines gives the following equation:

$\frac{sin(angleA)}{a} =\frac{sin(angleB)}{b} =\frac{sin(angleC)}{c}$

From the picture you presented, with triangle RST, side r is on the right and unlabeled, side s is shown in red, and side T is shown in green. The interior angles for R and S are unlabeled, and angle T is defined as 74 degrees. The Law of Sines would give the following relationship for your triangle:

$\frac{sin(angleR)}{r} =\frac{sin(angleS)}{s} =\frac{sin(angleT)}{t}$

Substituting known values...

$\frac{sin(angleR)}{r} =\frac{sin(angleS)}{12.7ft} =\frac{sin(74^{o} )}{12.3ft}$

It should be noted that without information about the length of side r, angle R cannot be found directly with the Law of Sines because that portion of the equation holds two unknowns. However, if the other two angles of the triangle are known, angle R can be solved for by using the Triangle Sum Theorem.

<h3><u>Finding Angle S</u></h3>

Focusing in on the ratio with known values on the far right, and the ratio containing angleS first:

$\frac{sin(angleS)}{12.7ft} =\frac{sin(74^{o} )}{12.3ft}$

... multiplying both sides of the equation by 12.7 ft...

$sin(angleS) =\frac{sin(74^{o} ) * 12.7ft}{12.3ft}$

Note that the right side of the equation has units of feet in both the numerator and denominator (with no addition or subtraction), so the units will cancel. Simplifying the right side of the equation by evaluating the expression in a calculator (make sure you're in "degree" mode), will yield a unit-less number:

$sin(angleS) = .9925222389...$

Undoing the sine function, and solving for the measure of angle S will require taking the "arcsin" of both sides of the equation...

$arcsin(sin(angleS))=arcsin(0.9925222389...)\\angleS=arcsin(0.9925222389...)\\angleS=82.98876661...^o\\angleS=83^o$ <u />

<u />

<h3><u>Finding Angle R</u></h3>

Knowing both angleS and angleT, we can apply the Triangle Sum Theorem to solve for angle R.

Triangle Sum Theorem: The sum of all 3 interior angles in a triangle (in a plane) is 180 degrees.

$angleR+angleS+angleT=180^o\\angleR+83^o+74^o=180^o$

Using the subtraction property of equality to isolate angleR and combining like terms...

$angleR=180^o-83^o-74^o\\angleR=23^o$

8 0

1 year ago