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3 years ago

Can someone help me with this asap please, thanks!

Mathematics

1 answer:

kvv77 [185]3 years ago

5 0

Answer:

Given: $r\perp s$ and $t\perp s$ .

Prove: $r\parallel t$

Since $r\perp s$ , By the definition of perpendicular lines angles 1, 2, 3 and 4 are 90 degree. similarly $t\perp s$ , it means angles 5, 6, 7 and 8 are 90 degree.

We can say that,

$\angle 1\cong \angle 5$ , $\angle 2\cong \angle 6$ , $\angle 3\cong \angle 7$ and $\angle 4\cong \angle 8$

From figure it is noticed that the angle 1 and 5 are corresponding angles.

If two parallel lines are intersected by a transversal, then the corresponding angles are equal.

Since corresponding angles are congruent, therefore the line must be parallel to each other.

Hence proved that $r\parallel t$ .

You might be interested in

If a=mg-kv2/m,find,correct to the nearest whole number the value of v when a=2.8,m=12,g=9.8 and k=8/3

Cerrena [4.2K]

Answer:

v = 23

Step-by-step explanation:

Formatting the question gives;

a = mg - kv² / m

Make v subject of the formula as follows;

(i) Multiply both sides by m

ma = m²g - kv²

(ii) Collect like terms

kv² = m²g - ma

(iii) Divide through by k

v² = (m²g - ma) / k

(iv) Take the square root of both sides

v = √ [(m²g - ma) / k] --------------(ii)

From the question:

a = 2.8

m = 12

g = 9.8

k = 8/3

Substitute these values into equation (i) as follows;

v = √ [(12²(9.8) - 12(2.8)) / (8/3)]

v = √ [(1411.2 - 33.6) / (8/3)]

v = √ [1377.6 / (8/3)]

v = √ [1377.6 x (3/8)]

v = √ [1377.6 x 3 / 8)]

v = √ [516.6]

v = 22.73

v = 23 [to the nearest whole number]

Therefore v = 23 to the nearest whole number

3 0

4 years ago

Solve: 10 2/3+2h≤4 1/3

PtichkaEL [24]

Answer:

$\large\boxed{h\leq-\dfrac{19}{6}\to h\in\left(-\infty,\ -\dfrac{19}{6}\right]}$

Step-by-step explanation:

$\text{First convert the mixed numbers to the improper fractions:}\\\\10\dfrac{2}{3}=\dfrac{10\cdot3+2}{3}=\dfrac{32}{3}\\\\4\dfrac{1}{3}=\dfrac{4\cdot3+1}{3}=\dfrac{13}{3}\\\\\text{Solve the inequality}\\\\10\dfrac{2}{3}+2h\leq4\dfrac{1}{3}\\\\\dfrac{32}{3}+2h\leq\dfrac{13}{3}\qquad\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\dfrac{32}{3\!\!\!\!\diagup_1}+3\cdot2h\leq3\!\!\!\!\diagup^1\cdot\dfrac{13}{3\!\!\!\!\diagup_1}\\\\32+6h\leq13\qquad\text{subtract 32 from both sides}\\\\6h\leq-19\qquad\text{divide both sides by 6}\\\\h\leq-\dfrac{19}{6}$