Ask question

Ask question

All categories

2 years ago

10

Let $M$, $N$, and $P$ be the midpoints of sides $\overline{TU}$, $\overline{US}$, and $\overline{ST}$ of triangle $STU$, respect

ively. Let $\overline{UZ}$ be an altitude of the triangle. If $\angle TSU = 71^\circ$, $\angle STU = 36^\circ$, and $\angle TUS = 73^\circ$, then what is $\angle NZM + \angle NPM$ in degrees?

1 answer:

AysviL [449]2 years ago

5 0

Answer:

<NPM + <NZM = 146°

Step-by-step explanation:

Given:

In triangle STU: M, N and P are the midpoints of the line TU, US and ST respectively.

Line UZ is the altitude of the triangle STU

<TSU =71°, <TSU = 36°, <TUS = 73°

From the diagram:

N is the midpoint of line SU and M is the mid point of line UT.

∴ Line NM is parallel to line ST

P is the mid point of line ST and M is the mid point of line UT

∴Line PM is parallel to line SU

N is the mid point of line SU and P is the mid point of line ST

∴Line NP is parallel to line UT

Δ SPN = Δ STU = 36°

<SPN + <NPM + <MPT (Sum of angles in a triangle = 180°)

<UST = <MPT = 71°

36° + <NPM + 71° = 180°

<NPM + 107° = 180°

<NPM= 180°-107°

<NPM= 73°

In ΔSNZ, line SN = line NZ

∴ side SN = side NZ

< NSZ = <NZS = 71°

<MTZ = <MZT = 36°

<NZS + <NZM <MZT = 180° (Sum of angles in a triangle = 180°)

71° + <NZM + 36° = 180°

107° + <NZM = 180°

<NZM = 180° - 107°

<NZM = 73°

<NPM + <NZM =73° + 73°

<NPM + <NZM = 146°

You might be interested in

Adding and subtracting positive and negative numbers

algol13

Answer:

Step-by-step explanation:

Rule Example

+(+) Two like signs become a positive sign 3+(+2) = 3 + 2 = 5

−(−) 6−(−3) = 6 + 3 = 9

+(−) Two unlike signs become a negative sign 7+(−2) = 7 − 2 = 5

−(+) 8−(+2) = 8 − 2 = 6

5 0

2 years ago

Read 2 more answers

Help: theoretical and experimental probability

Natalija [7]

D is the right one, in all these suggestions , the number cube is rolled 1 time.

5 0

3 years ago

Jimmy has decided to limit his spending to $175 per day. His roundtrip airfare is $1750. How many days can he travel if he can s

Sever21 [200]

Hi

The answer is 22 days .

If we divide 4,000 by 175 , we get 22.8 , which is basically 23 , but if we multiply 23 by 175 we get 4,025 .

22 times 175= 3850
3850 < 4000

Answer: 22

4 0

3 years ago

Point A is the incenter of ΔDEF. Point A is the incenter of triangle D E F. Lines are drawn from the points of the triangle to p

Ksenya-84 [330]

Answer:

The correct option is;

11

Step-by-step explanation:

The incenter is the point where the interior angle bisectors of the triangle meet, hence the interior angles are each bisected into two by the segments DA, EA, and FA respectively

Given that point A is the incenter of ΔDEF, we have;

∠FDE = ∠XDY is bisected by DA which gives ∠XDA = ∠YDA

∠LDA = ∠XDA = 27

∠MDA = ∠YDA = 3··n - 6

Therefore, 27 = 3·n - 6

3·n = 27 + 6 = 33

n = 33/3 = 11°

3 0

2 years ago

I need help from question 11- 16! Please help!

insens350 [35]

<h2>11. Find discriminant.</h2>

Answer: D) 0, one real solution

A quadratic function is given of the form:

$ax^2+bx+c=$

We can find the roots of this equation using the quadratic formula:

$x_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Where $\Delta=b^2-4ac$ is named the discriminant. This gives us information about the roots without computing them. So, arranging our equation we have:

$4a^2-4a-6=-7 \\ \\ Adding \ 7 \ to \ both \ sides \ of \ the \ equation: \\ \\ 4a^2-4a-6+7=-7+7 \\ \\ 4a^2-4a+1=0 \\ \\ Then \ the \ discriminant: \\ \\ \Delta=(-4)^2-4(4)(1) \\ \\ \Delta=16-16 \\ \\ \boxed{Delta=0}$

<em>Since the discriminant equals zero, then we just have one real solution.</em>

<h2>12. Find discriminant.</h2>

Answer: D) -220, no real solution

In this exercise, we have the following equation:

$-r^2-2r+14=-8r^2+6$

So we need to arrange this equation in the form:

$ax^2+bx+c=$

Thus:

$-r^2-2r+14=-8r^2+6 \\ \\ Adding \ 8r^2 \ to \ both \ sides \ of \ the \ equation: \\ \\ -r^2-2r+14+8r^2=-8r^2+6+8r^2 \\ \\ Associative \ Property: \\ \\ (-r^2+8r^2)-2r+14=(-8r^2+8r^2)+6 \\ \\ 7r^2-2r+14=6 \\ \\ Subtracting \ 6 \ from \ both \ sides: \\ \\ 7r^2-2r+14-6=6-6 \\ \\ 7r^2-2r+8=0$

So the discriminant is:

$\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}$

<em>Since the discriminant is less than one, then there is no any real solution</em>

<h2>13. Value that completes the squares</h2>

Answer: C) 144

What we need to find is the value of $c$ such that:

$x^2+24x+c=0$

is a perfect square trinomial, that are given of the form:

$a^2x^2\pm 2axb+b^2$

and can be expressed in squared-binomial form as:

$(ax\pm b)^2$

So we can write our quadratic equation as follows:

$x^2+2(12)x+c \\ \\ So: \\ \\ a=1 \\ \\ b=12 \\ \\ c=b^2 \therefore c=12^2 \therefore \boxed{c=144}$

Finally, the value of $c$ that completes the square is 144 because:

$x^2+24x+144=(x+12)^2$

<h2>14. Value that completes the square.</h2>

Answer: C) $\frac{121}{4}$

What we need to find is the value of $c$ such that:

$z^2+11z+c=0$

So we can write our quadratic equation as follows:

$z^2+2\frac{11}{2}z+c \\ \\ So: \\ \\ a=1 \\ \\ b=\frac{11}{2} \\ \\ c=b^2 \therefore c=\left(\frac{11}{2}\left)^2 \therefore \boxed{c=\frac{121}{4}}$

Finally, the value of $c$ that completes the square is $\frac{121}{4}$ because:

$z^2+11z+\frac{121}{4}=(x+\frac{11}{2})^2$

<h2 /><h2>15. Rectangle.</h2>

In this problem, we need to find the length and width of a rectangle. We are given the area of the rectangle, which is 45 square inches. We know that the formula of the area of a rectangle is:

$A=L\times W$

From the statement we know that the length of the rectangle is is one inch less than twice the width, this can be written as:

$L=2W-1$

So we can introduce this into the equation of the area, hence:

$A=L\times W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length$

$A=(2W-1)(W) \\ \\ But \ A=45: \\ \\ 45=(2W-1)(W) \\ \\ Distributive \ Property:\\ \\ 45=2W^2-W \\ \\ 2W^2-W-45=0 \\ \\ Quadratic \ Formula: \\ \\ x_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ W_{1}=\frac{-(-1)+ \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{1}=\frac{1+ \sqrt{1+360}}{4} \therefore W_{1}=5 \\ \\ W_{2}=\frac{-(-1)- \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{2}=\frac{1- \sqrt{1+360}}{4} \therefore W_{2}=-\frac{9}{2}$

The only valid option is $W_{1}$ because is greater than zero. Recall that we can't have a negative value of the width. For the length we have:

$L=2(5)-1 \\ \\ L=9$

Finally:

$The \ length \ is \ 9 \ inches \\ \\ The \ width \ is \ 5 \ inches$

<h2>16. Satellite</h2>

The distance in miles between mars and a satellite is given by the equation:

$d=-9t^2+776$

where $t$ is the number of hours it has fallen. So we need to find when the satellite will be 452 miles away from mars, that is, $d=452$ :

$d=-9t^2+776 \\ \\ 452=-9t^2+776 \\ \\ 9t^2=776-452 \\ \\ 9t^2=324 \\ \\ t^2=\frac{324}{9} \\ \\ t^2=36 \\ \\ t=\sqrt{36} \\ \\ \boxed{t=6h}$

Finally, <em>the satellite will be 452 miles away from mars in 6 hours.</em>

3 0

3 years ago