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Radda [10]
3 years ago
7

Simplify the expression below. 14a8y3 − 7a4y5 + 28a12y2 7a4y

Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0

Answer:

14a^{8}y^{3} - 7a^{4}y^{5} + 4a^{8} y

Step-by-step explanation:

14a^{8}y^{3} - 7a^{4}y^{5} + \frac{28a^{12}y^{2}}{7a^{4} y}

14a^{8}y^{3} - 7a^{4}y^{5} + 4a^{8} y

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Find the length of the missing side.<br> 50 m<br> 30 m<br> The length of the missing side is<br> m.
satela [25.4K]

Answer: The missing length is 40 m.

Step-by-step explanation:

30^2 + b^2 = 50^2

900 + b^2 = 2500

-900              -900

       b^2 = 1600

b=  40

3 0
3 years ago
During one term Rachel took 2 mathematics test her mean score was 75 she scored 12 points more on the first test then the second
MaRussiya [10]

Answer: she scored 69 points on the second test.

Step-by-step explanation:

Let x represent the number of points that she scored in the first test.

Let y represent the number of points that she scored in the second test.

If her mean score was 75, it means that

(x + y)/2 = 75

x + y = 75 × 2

x + y = 150- - - - - - - - - 1

If she scored 12 points more on the first test than the second test, it means that

x = y + 12

Substituting x = y + 12 into equation 1, it becomes

y + 12 + y = 150

2y + 12 = 150

2y = 150 - 12

2y = 138

y = 138/2

y = 69

3 0
3 years ago
Find the area of the shaded region
zepelin [54]

Answer:

area of a circle πr^2

r = 10/2 r = 5

π5^2 = 78.5

area of the square length X width

7 x 7 = 49

78.5 - 49 = 29.5

shaded region = 29.5 square cm

3 0
3 years ago
Read 2 more answers
A right triangle has side lengths 7, 24, and 25 as shown below.
Alla [95]

9514 1404 393

Answer:

  • tan(A) = 7/24
  • sin(A) = 7/25
  • cos(A) = 24/25

Step-by-step explanation:

The mnemonic SOH CAH TOA can help you remember the trig ratios:

  Sin = Opposite/Hypotenuse

  Cos = Adjacent/Hypotenuse

  Tan = Opposite/Adjacent

With respect to angle A, the sides are Adjacent = 24; Opposite = 7, Hypotenuse = 25. Then the desired trig ratios are ...

  tan(A) = 7/24

  sin(A) = 7/25

  cos(A) = 24/25

5 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

3 0
3 years ago
Read 2 more answers
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