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3 years ago

I need help Idk this one

Mathematics

1 answer:

mihalych1998 [28]3 years ago

6 0

Answer:

1.6

Step-by-step explanation:

First things first is to find what line QP corresponds with which is line MN

After that you find what multiplied into 5 would make it 2. The easiest way of ding so is to divide 2 by 5 which is 0.4. Now you want to find what line PR corresponds with which is OM. Now you multiply 4 (length of line OM) by 0.4 to find the length of line PR which is x. So now line PR is equal to 1.6 and x. Which also means that x=1.6

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Hi help please and thank you

Charra [1.4K]

Answer:

looks hard

Step-by-step explanation:

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3 years ago

Can you help me with this question in geometry?

Svetllana [295]

I believe this just uses the Pythagorean Theorem. A^2+B^2=C^2. So:
15^2+B^2=25^2. Simplified this is,
225+B^2=625
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B^2=400
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8 0

3 years ago

The linear inequality y<-2/3x+2 is graphed. Determine a solution for the inequality. A) (0,2) B) (3,0) C) (0,0) D) (0,3)

jenyasd209 [6]

Hello!

The graph y < -2/3x + 2 has a dotted line. That means that any points on the dotted line or not shaded, is not a solution to this inequality. Since we are given 4 choices, we can substitute those values into the given inequality to see if it is true.

A) (0, 2)

2 < -2/3(0) + 2

2 < 2, this is false.

B) (3, 0)

0 < -2/3(3) + 2

0 < -2 + 2

0 < 0, this is false.

C) (0, 0)

0 < -2/3(0) + 2

0 < 2, this is true.

D). (0, 3)

3 < -2/3(0) + 2

3 < 2, this is false.

To check if choice C) (0, 0) is true, we should look at the given graph.

Since (0, 0) is in the shaded area, and is not graphed on the dotted line, therefore, a solution to this linear inequality is C, (0, 0).

5 0

4 years ago

Plz hurry!!!! thank you!!!!

Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = $\frac{115}{2}$ = 57.5°

Now, tan(57.5°) = $\frac{OS}{SN}$

⇒ 1.5697 = $\frac{R}{SN}$

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = $\frac{NQ}{QM}$

⇒ tan65° = $\frac{NQ}{QM}$

⇒ QM = $\frac{2R}{2.1445}$

QM = 0.9326 R .

⇒ MT = MQ + QT

= 0.9326 R + 0.637 R (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = ( $\frac{NP + MK}{2}$ ) × (ST)

= ( $\frac{1.274 R + 3.1392 R}{2}$ ) × 2 R

= 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

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3 years ago

Write a word phrase that can be represented by b – 6.

Aleonysh [2.5K]

D , the answer is "six less than a number b"

4 0

3 years ago

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