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3 years ago

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

3 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