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3 years ago

Solve the following problem: –25 + 37 =

Mathematics

2 answers:

Kay [80]3 years ago

8 0

here all the answers

Step-by-step explanation

1. Solve the following problem: –25 + 37 =

(0 pts) 62

(0 pts) –62

(1 pt) 12

(0 pts) –1

2. Solve the following problem: 17 + 50 – 100 =

(1 pt) –33

(0 pts) 33

(0 pts) 167

(0 pts) –167

3. Rewrite the following expression using the Distributive Property. 5(2x – 8)

(0 pts) –30x

(1 pt) 10x – 40

(0 pts) 10x + 40

(0 pts) 10x – 8

4.Simplify the following expression: 3(5 – 12) + 4 ÷ 2

(1 pt) –19

(0 pts) 19

(0 pts) –23

(0 pts) –8.5

Simplify the following expression: -16-8/2 + 25 ÷ 5 + 1

(0 pts) –18

(0 pts) 6

(0 pts) 10

(1 pt) –6

10/10 100%

Lerok [7]3 years ago

3 0

1. 12 is your answer

2. -33 is your answer

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f(40) = log2 40

40 = 2^y

2^5 = 32 and 2^6 = 64

so f(40) lies between integers 5 and 6.

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Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Now given a rope hanging from the top of a pole. The end lying on the ground is 3 chi long. When tightly stretchedit is 8 chi fr

drek231 [11]

Answer:

Height of the pole = 9.17 chi.

Length of the rope = 12.17 chi.

Step-by-step explanation:

Given that the rope is hanging from the top of a pole having height x chi and the portion of the rope lying on the ground is 3 chi.

So, the length of the rope= x + 3 chi.

Let AB represents the pole in the figure, and one end of the rope is at point A.

When the rope is tightly stretched, let C be the other end of the rope as shown in the triangle.

The length of the rope = AC.

\Rightarrow AC=x+3 chi.

Distance from the bottom of the pole, point A, to the other end of the pole, point B, is 8 chi.

So, BC=8 chi.

As the triangle ABC is a right-angled triangle, so by using Pythagoras theorem,

$AC^2= AB^2+BC^2$

$\Rightarrow (x+3)^2=x^2+8^2$

$\Rightarrow x^2+6x+9=x^2+64$

$\Rightarrow 6x=64-9=55$

$\Rightarrow x=55/6=9.17$ chi.

Hence, the height of the pole, $AB=x=9.17$ chi,

and the length of the rope, $x+3=9.17+3=12.17$ chi.

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