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3 years ago

Helpp me will give brainliest!!!!

Mathematics

1 answer:

olga_2 [115]3 years ago

5 0

Let's find area of the base.: 5x4.3=21.5, 21.5/2=10.75

Area of the Faces : 5x3=15/2=7.5x4=30

Let's add 10.75 +30=40.75

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Diego is solving the inequality 100 − 3x ≥ -50. He solves the equation 100 − 3x = -50 and gets x = 50. What is the solution to t

ZanzabumX [31]

Answer:

x≤50

Step-by-step explanation:

100-3x≥-50

subtracting 100 at both side on inequality

-100+100-3x≥--50-100

-3x≥-150

Dividing both sides by 3

-3x/3≥-150/3

-x≥--50

Adding 50 on both sides

-x+50≥-50+50

-x+50≥0

Adding x on both sides

-x+x+50≥0+x

50≥x

so,

x≤50

7 0

3 years ago

Read 2 more answers

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

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