Ask question

Ask question

All categories

2 years ago

5

explain how reading the digits ro the right of the decimal point and knowing the name of the least place value help yoi read a d

ecimal number.use examples .604 and 1.604 in your explanation

1 answer:

AVprozaik [17]2 years ago

7 0

Chicken sandwich with my baby and my grandma is a little too much to do and I wanna was a way better then you got a baby and stuff and I wanna just like a

You might be interested in

Who could help me on the last part

anyanavicka [17]

Step-by-step explanation:

X^a times X^b = X^a + b

so 4x times 6x = 24x^(1+1)

6 0

2 years ago

Write 425% as a fraction or mixed number in simplest form.

Temka [501]

Answer:

17/40

here! hope this helps!

8 0

2 years ago

Simplify (8^4/10^4)^-1/4using rational exponents

Vlad1618 [11]

Answer:

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

Step-by-step explanation:

<u><em>Explanation</em></u>

Given

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$

By using algebra formulas

$(\frac{a}{b} )^{m} = \frac{a^m}{b^n}$

$(\frac{8^4}{10^4} )^{-\frac{1}{4} } = \frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } }$

$(a^{m} )^{n} = a^{mn}$

$\frac{(8^{4})^{\frac{-1}{4} } }{(10^{4} )^{\frac{-1}{4} } } = \frac{8^{4X\frac{-1}{4} } }{10^{4 X\frac{-1}{4} } }$

= $\frac{8^{-1} }{10^{-1} }$

$= \frac{10}{8} \\= \frac{5}{4}$

<u><em>Final answer:-</em></u>

$(\frac{8^4}{10^4} )^{-\frac{1}{4} }$ = $\frac{5}{4}$

7 0

3 years ago

What does x equal (x/4)+12=38

user100 [1]

$\frac{x}{4} +12=38\\\\ \frac{x}{4}=26\\\\x=104$

8 0

3 years ago

Read 2 more answers

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago