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djyliett [7]
3 years ago
6

Consider the plane, T, and three collinear points (but not coplanar) points P, Q, and W.

Mathematics
1 answer:
Montano1993 [528]3 years ago
5 0

Answer:

vvv dvw

Step-by-step explanation:

dnbvfv nb f

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Unit 4 Test, Part 2: Congruence and Constructions<br> Help with my math please
lakkis [162]

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

\overline{NM} = 20

\overline{MK} = 15

\overline{KN} = 25

\overline{TR} = 3x - 1

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to \overline{TR} is \overline{NM}. Thus:

\overline{TR} ≅ \overline{NM}

c. Since \overline{TR} ≅ \overline{NM}, therefore:

\overline{TR} = \overline{NM}

3x - 1 = 20 (substitution)

Add 1 to both sides

3x = 20 + 1

3x = 21

Divide both sides by 3

x = \frac{21}{3}

x = 7

2. a. Slope of LK = \frac{rise}{run} = \frac{4}{3}

Slope of LM = \frac{rise}{run} = -\frac{3}{5}

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

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

Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}

Lk = \sqrt{(3)^2 + (4)^2}

Lk = \sqrt{9 + 16}

Lk = \sqrt{25}

Lk = 5

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

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

LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}

LM = \sqrt{(5)^2 + (-3)^2}

LM = \sqrt{25 + 9}

LM = \sqrt{34}

LM = 5.8 (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.

8 0
3 years ago
What is the area of this figure?
umka21 [38]
137.5 I took the test and got 100% 


3 0
2 years ago
Read 2 more answers
PLEASE HELP 5 STAR + BRAINLY IF RIGHT!
Sever21 [200]
Right your ratio of \frac{8}{12}
Simplify fraction to 3/4
Now set it up with c/100 to find the percent (c equaling the percentage of caremel brownies)

Cross multiplty:
\frac{2}{3} \frac{c}{100}

Crossmultiply:
3c = 200
Divide (solve equation)

c = <span>66.6666666667</span>
7 0
3 years ago
Read 2 more answers
Please help thank you very much
-BARSIC- [3]
Hello!

The answer is 34°

Hope this helps!
7 0
3 years ago
From a circular cylinder of diameter 10 cm and height 12 cm are conical cavity of the same base radius and of the same height is
Nataliya [291]
<h3>Volume of the remaining solid = 628 cm^2</h3>

<h3>Whole surface area = 659.4 cm^2</h3>

Step-by-step explanation:

Now, Given that:-

Diameter (d) = 10 cm

So, Radius (r) = 10/2 = 5cm

Height of the cylinder = 12cm.

volume \: of \: the \: cylinder \:  =  \pi {r}^{2} h

=  > \pi \times  {5}^{2} \times  12 {cm}^{3}   = 300\pi {cm}^{3}

Radius of the cone = 5 cm.

Height of the cone = 12 cm.

slant \: height \: of \: the \: cone \:  =  \sqrt{ {h}^{2}  + \:  {r}^{2} }

=  >  \sqrt{ {5}^{2}+{12}^{2} } cm \:  = 13cm

Volume of the cone = 1/3 *πr^2h

=  >  \frac{1}{3} \pi \times  {5}^{2}   \times 12 {cm}^{3}  = 100\pi {cm}^{3}

therefore, the volume of the remaining solid

= 300\pi {cm}^{3}  - 100\pi {cm}^{3}  \\  = 200 \times 3.14 {cm}^{3}  = 628 {cm}^{3}

Curved surface of the cylinder =

2\pi \: rh \:  = 2\pi \times 5 \times 12 {cm}^{2}  \\  = 120\pi {cm}^{2} .

curved \: surface \: of \: the \: cone \:  = \pi \: rl \\  = \pi \times 5 \times 13 {cm}^{2}  \\  = 65\pi {cm }^{2} \\ area \: of \: (upper)circular \: base \: \\  of \: cylinder \:  =  \\ =  \pi \:  {r}^{2}  = \pi \times  {5}^{2}

therefore, The whole surface area of the remaining solid

= curved surface area of cylinder + curved surface area of cone + area of (upper) circular base of cylinder

= 120\pi {cm}^{2}  + 65\pi {cm }^{2}  + 25 \pi {cm}^{2}  \\  = 210 \times 3.14 {cm}^{2}  = 659.4 {cm}^{2}

<h3>Hope it helps you!!</h3>

6 0
2 years ago
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