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3 years ago

-256÷457.6 a -16 b -1.6c 1.6 d 16

Mathematics

2 answers:

Aleksandr [31]3 years ago

7 0

Answer the answer is c

Step-by-step explanation: one word calculator

sp2606 [1]3 years ago

4 0

Answer:

Step-by-step explanation:

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vekshin1

B) 6, 8, 10

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8 0

3 years ago

(ASAP)<br> Solye for n.<br> 24 = 3(n – 5) <br> n=

Paul [167]

Answer: can u check out the image cuz the answer is in the photo and hope this helps can u give me brainliest

Step-by-step explanation:

24=3(n-5)

Distribute.

24=3(n) + 3(-5)

24 = 3n -15

Add 15 to both sides.

39 = 3n

Divide both sides by 3.

n = 13.

I hope this helps!.

6 0

3 years ago

A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 19 cm long and 4 cm wide. The di

alexira [117]

Answer:

Since we have two semi-circles we can say that we have a full circle. using the circle area formula we can say that

3.14*2^2 = 12.56 cm

Using the rectangle area formula we can say that

19*4 = 76 cm

Adding them both together is

12.56 + 76 = 88.56 cm

So the bookmark has an area of 88.56cm^2

8 0

3 years ago

What is the surface area of a cube whose edges are 11 inches long? A. 66 square inches B. 660 square inches C. 726 square inches

Alex787 [66]

Answer:

C. 726 in^2.

Step-by-step explanation:

Each face has area 11^2 = 121 in^2 ( because each face is a square).

There are 6 faces so the total surface area = 6 * 121

= 726 in^2.

3 0

3 years ago

Read 2 more answers

7. By using binomial expansion show that the value of (1.01)^12 exceed the value of (1.02)^6 by 0.0007 correct to four decimal p

BlackZzzverrR [31]

Binomial expansion is used to factor expressions that can be expressed as the power of the sum of two numbers.

The proof that (1.01)^12 exceeds (1.02)^6 by 0.0007 is $\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }$

The expressions are given as:

$\mathbf{(1.01)^{12}\ and\ (1.02)^6}$

A binomial expression is represented as:

$\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}$

Express 1.01 as 1 + 0.01

So, we have:

$\mathbf{(1.01)^{12} = (1 + 0.01)^{12}}$

Apply the above formula

$\mathbf{(1.01)^{12} = ^{12}C_0 \times 1^{12 - 0} \times 0.01^0 + ......... .......... + ^{12}C_{12} \times 1^{12 - 12} \times 0.01^{12} }}$

$\mathbf{(1.01)^{12} = 1 \times 1 \times 1 + ......... .......... + 1 \times 1 \times 10^{-24} }}$

$\mathbf{(1.01)^{12} = 1 + ......... .......... + 10^{-24} }}$

This gives

$\mathbf{(1.01)^{12} = 1.1268\ (approximated)}$

Similarly,

Express 1.02 as 1 + 0.02

So, we have:

$\mathbf{(1.02)^6 = (1 + 0.02)^6}$

Apply $\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}$

$\mathbf{(1.02)^6 = ^6C_0 \times 1^{6 - 0} \times 0.02^0 + ^6C_1 \times 1^{6 - 1} \times 0.02^1 +.............. + ^6C_6 \times 1^{6 - 6} \times 0.02^6 }$ $\mathbf{(1.02)^6 = 1 \times 1 \times 1 + 6 \times 1 \times 0.02 +.............. + 1 \times 1 \times 6.4 \times 10^{-11} }$