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Marta_Voda [28]

3 years ago

5

Triangle F G E is shown. Angle F E G is a right angle. The length of hypotenuse F G is 15.2 meters and the length of F E is 9 me

ters. Angle F G E is x.
Which equation can be used to find the measure of angle FGE?

2 answers:

kkurt [141]3 years ago

8 0

The anser is on that guys link bbg

cluponka [151]3 years ago

3 0

Answer:

B. 40.2 degrees

Step-by-step explanation:

Consider ΔFGE. Given that ∠FEG = 90° and ∠FGE is x°.

To find the value of ∠FGE use trigonometric ratios.

Use sine ratio to find the value of ∠FGE.

Rewriting it in terms of x° as follows,

Now find the opposite and hypotenuse side of the angle ∠FGE.

So opposite side of ∠FGE is FE and hypotenuse side of ∠FGE is FG.

Substituting the values,

Given that length of FE is 12 inch and FG is 18.6 inch.

Substituting the value,

To find the value of x, taking inverse sin.

Calculating the value,

Rounding to nearest tenth the value of angle is 40.2

Therefore measure of ∠FGE is 40.2° .

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Find the equation of the line.

NARA [144]

Answer:

y=4x+9

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5 0

4 years ago

Statement Reason m∠1 + m∠2 = 180° Angles 1 and 2 are supplementary angles. m∠2 + m∠3 = 180° Angles 2 and 3 are supplementary ang

mel-nik [20]

Angle 1 is congruent with angle3.

Good luck!!!

8 0

4 years ago

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

nika2105 [10]

Answer:

The components of $\vec{b}$ parallel and perpendicular to $\vec {a}$ are $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ and $\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$ , respectively.

Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$ , the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$ , dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$ , whose value is determined by Pythagorean Theorem.

The component of $\vec{b}$ parallel to $\vec {a}$ is:

$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

$\|\vec {a}\| = \sqrt{10}$

$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

$\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j$

$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

If $\vec b = 2\,i+j-3\,k$ and $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ , then:

$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$

4 0

3 years ago

Calculate the value of 4321 (base 5) + 343 (base 5)

AnnZ [28]

In the "1s" place,

1₅ + 3₅ = 4₅

In the "5s" place,

2₅ + 4₅ = (6)₅ = (5 + 1)₅ = 10₅ + 1₅ = 11₅

In the "25s" place, we carry the leading 1 from the "5s" place,

3₅ + 3₅ + 1₅ = (7)₅ = (5 + 2)₅ = 10₅ + 2₅ = 12₅

In the "125s" place, we carry over the 1 again,

4₅ + 1₅ = (5)₅ = (5 + 0)₅ = 10₅ + 0₅ = 10₅

Putting everything together,

4321₅ + 343₅ = 10214₅

8 0

3 years ago

Suppose you can afford only $200 a month in car payments and your best loan option is a 60-month loan at 3%. How much money coul

g100num [7]

Answer:

$11,130.47

Step-by-step explanation:

The amortization formula can be used. It tells you the monthly payment amount A for some principal P, interest rate r, and n payments.

A = P(r/12)/(1 -(1 +r/12)^(-n))

Filling in your values, we get ...

200 = P(.03/12)/(1 -(1 +.03/12)^-60) = P(.0025)/(1 -1.0025^-60)

P = 200(1 -1.0025^-60)/.0025 ≈ 200×55.6523577

P ≈ 11,130.47

The present value of the loan is $11,130.47.

8 0

4 years ago