All categories

3 years ago

1 MINUTE!!!

Mathematics

2 answers:

iris [78.8K]3 years ago

6 0

Answer:

About 9.38

Step-by-step explanation:

Since the $\sqrt{88}$ is between the $\sqrt{81}$ and the $\sqrt{100}$ which simplified is 9 and 10, you just need to guess and check. Eventually you'll get about 9.38 (I used Google cause I was lazy) And if it isn't between the numbers 9 & 10, it's wrong.

aleksley [76]3 years ago

5 0

Answer:

The answer is 2√22.

Step-by-step explanation:

You have to elaborate it by finding the common factors :

88 = 1×88

88 = 2×44

88 = 4×22

88 = 8×11

Next, you have to find the perfect square. In the factors above, 4 is the perfect square then, you can square root it :

$\sqrt{88}$

$= \sqrt{4 \times 22}$

$= \sqrt{4} \times \sqrt{22}$

$= 2 \times \sqrt{22}$

$= 2 \sqrt{22}$

You might be interested in

PQRS is a quadrilateral.

olga nikolaevna [1]

Answer:

y = 95° (being corresponding angle)

it is very easy.

you can also do

7 0

3 years ago

Read 2 more answers

What is the volume of this composite solid?

Marta_Voda [28]

Answer:

A: 64 cubic meters

Step-by-step explanation:

Multiply 7m, 2m, and 4m. You get 56

Then for the smaller cube multiply 2, 2, and 2

3 0

2 years ago

Read 2 more answers

How do you change 20.25 into a fraction

bearhunter [10]

Answer:

81/4

Step-by-step explanation:

I know this because I did this before

7 0

3 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$