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2 years ago

help me with one ty

Mathematics

1 answer:

trasher [3.6K]2 years ago

7 0

Answer:

Assuming that the angles of JKL and NOM are in equal order, we know that congruent angles in a congruent triangle are equal to the corresponding angles in the other triangle. And assuming that JKL has the corresponding angles in the same order as NOM, statement D is correct.

This is because KL would be equal to OM not NM, JK is equal to NO not OM, JK and KL are apart of the same triangle, and LJ does equal MN assuming the corresponding angles are in the same order.

Step-by-step explanation:

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The perimeter of a college basketball court is 96 meters and the length is 14 meters more than the width. What are the dimension

soldi70 [24.7K]

Answer:

length = 32m, width= 16m

Step-by-step explanation:

l=2w

2(2w+w)=96

2(3w)=96

6w=96,w= 96/6 = 16m

l=2w=2x16=32m

8 0

2 years ago

Read 2 more answers

Given: -16 = m/5 -18; Prove: m=10

spayn [35]

Answer:

I have proven m is 10. have a good day and give brainliest if possible.

3 0

2 years ago

In the given diagram if AB || CD, ∠ABO = 118° , ∠BOD = 152° , then find the value of ∠ODC.

GaryK [48]

Answer:

90°

Step-by-step explanation:

Through point O draw a ray on left side of O which is || to AB & CD and take any point P on it.

Therefore,

∠ABO + ∠BOP = 180° (by interior angle Postulate)

118° + ∠BOP = 180°

∠BOP = 180° - 118°

∠BOP = 62°.... (1)

Since, ∠BOP + ∠POD = ∠BOD

Therefore, 62° + ∠POD = 152°

∠POD = 152° - 62°

∠POD = 90°.....(2)

∠POD + ∠ODC = 180° (by interior angle Postulate)

90° + ∠ODC = 180°

∠ODC = 180° - 90°

$\huge\red {\boxed {m\angle ODC = 90°}}$

6 0

3 years ago

Find, to four significant digits, the distance between the graphs of 2x-3y+4=0 and 2x-3y+15=0.

shepuryov [24]

In analytical geometry, we can find the linear distance between two parallel lines by determining the coefficients of the variables and the constants. The general equation for a linear equation is: Ax + By = C.

Line 1: 2x - 3y = -4
Line 2: 2x - 3y = -15

In this case, A=2 and B=-3. The constants are C₁ = -4 and C₂ = -15. The distance follows this formula:

d = |C₁ - C₂|/(√(A²+B²)
d = |⁻4 - ⁻15|/(√(2²+⁻3²)
d = 3.051 units

4 0

3 years ago

Find the line integral with respect to arc length ∫C(9x+5y)ds, where C is the line segment in the xy-plane with endpoints P=(2,0

Gemiola [76]

(a) You can parameterize C by the vector function

r(t) = (x(t), y(t) ) = P (1 - t ) + Q t = (2 - 2t, 7t )

where 0 ≤ t ≤ 1.

(b) From the above parameterization, we have

r'(t) = (-2, 7) ==> ||r'(t)|| = √((-2)² + 7²) = √53

Then

ds = √53 dt

and the line integral is

$\displaystyle\int_C(9x(t)+5y(t))\,\mathrm ds = \boxed{\sqrt{53}\int_0^1(17t+18)\,\mathrm dt}$

$\displaystyle\sqrt{53}\int_0^1(17t+18)\,\mathrm dt = \sqrt{53}\left(\frac{17}2t^2+18t\right)\bigg|_{t=0}^{t=1} = \boxed{\frac{53^{3/2}}2}$

6 0

2 years ago

￼help me with one ty

help me with one ty