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2 years ago

Please need help asap!

Mathematics

1 answer:

beks73 [17]2 years ago

6 0

Answer:

Step-by-step explanation:

(z-33)/z = 1/2

2z-66=z

z=66

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89.396668... rational or irrational number?

aksik [14]

Answer:

89.396668... is a rational number

4 0

2 years ago

State true or false. If 9 is the solution of variable x in the equation fraction numerator 5 straight x space minus space 7 spac

Gekata [30.6K]

Answer:

False

Step-by-step explanation:

Based on what is stated in the question :

Expressing the worded equation Mathematically,

(5x - 7) / 2 = y

If y = 28

(5x - 7) / 2 = 28

Cross multiply

5x - 7 = 28 * 2

5x - 7 = 56

5x = 56 + 7

5x = 63

x = 63 / 5

x = 12.6

Hence, x ≠ 9

8 0

2 years ago

Two planes leave an airport at the same time, one flying east, the other flying west. The eastbound plane travels 130 mph slow

Art [367]

Answer:

Speed of the slower plane = 320 mph

Speed of the faster plane = 450 mph

Step-by-step explanation:

Let the speed of the slower plane = x mph

And the speed of the faster plane = y mph

Difference in their speeds = 130 mph

y - x = 130 --------(1)

Formula for the distance traveled = Speed × time

Distance traveled by the faster plane in 5 hours = 5y miles

Distance traveled by the slower plane in 5 hours = 5x miles

Total distance traveled by both the planes in 5 hours = 5(x + y) miles

5(x + y) = 3850

x + y = 770 ---------(2)

By adding equations (1) and (2)

2y = 900

y = 450 mph

From equation (2)

x = 770 - 450

x = 320 mph

6 0

2 years ago

The scale of map is 1cm : 70km. What is the actual distance between towns that are 5cm apart on the map

Illusion [34]

Answer: 350km

Step-by-step explanation:

Since we are informed that the scale of map is 1cm : 70km. Therefore, the actual distance between towns that are 5cm apart on the map will be:

= 5 × 70km

= 350km

Therefore, the answer is 350km

3 0

2 years ago

Use mathematical induction to prove

Alex17521 [72]

$Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\$

Formula works when n=1

Assume the formula also works, when n=k.

Prove that the formula works, when n=k+1

$1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\$

Since the formula has been proven with n=1 and n=k+1, it is true. $\square$

7 0

1 year ago