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2 years ago

Wolf brothers help me out, please no link

Mathematics

2 answers:

soldi70 [24.7K]2 years ago

6 0

Answer:

21 cards left to be signed

Step-by-step explanation:

1/3 of 72 is 24

So Harrison signed 24

3/8 of 72 is 27

Sp Alwine signed 27

24+27=51

So 72-51=21

21 cards left to be signed

Rudik [331]2 years ago

6 0

Answer:

21 cards left

Step-by-step explanation:

72 - (Harrison + Alwine) = Cards left to sign
Harison = 1/3 x 72
Alwine = 3/8 x 72

<u>Solution:</u>

72 - {(1/3 x 72) + (3/8 x 72)} = Cards left to sign
=> 72 - {24 + 27} = Cards left to sign
=> 72 - {51} = Cards left to sign
=> 72 - 51 = Cards left to sign
=> 21 cards left

Hence, 21 cards are left when Harrison signs 1/3 of the cards and Alwine signs 3/8.

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What is the slope of the line that passes through the points (-2,3) and (-1,8)

yaroslaw [1]

Answer:

Slope m = 5

Step-by-step explanation: sorry if Im wrong but If i am then it is just 5:)

5 0

2 years ago

Solve for x. And graph 7(3x+4)+2<156

blsea [12.9K]

$7(3 \times x + 4) + 2 < 156 \\ 21 \times x + 28 + 2 < 156 \\ 21 \times x + 30 < 156 \\ 21 \times x < 156 - 30 \\ 21 \times x < 126 \\ x < \frac{126}{21} \\ x < 6$

8 0

3 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$