All categories

3 years ago

QUICK MATH QUESTION!! :)

Mathematics

2 answers:

Marianna [84]3 years ago

4 0

It will be .0866 since a negative power will result in the number being smaller.

Alex3 years ago

3 0

You should move the decimal left when it's a negative exponent. Moving the decimal 2 digits left you get 0.08866

You might be interested in

Kai lives 3 kilometers from school, Brea lives 30,000 meters from school, and Cal lives 30,000 centimeters from school. Which st

Sedbober [7]

Answer:

D. Brea lives the farthest from school

Step-by-step explanation:

The easiest way to go about this is to get them all to the same unit. Let's do meters :)

3 kilometers = 3,000 meters (Kai)

30,000 centimeters = 300 meters (Cal)

and then of course 30,000 meters is already in the correct unit (Brea)

Given all of this, you can easily tell that the answer is D

6 0

3 years ago

34.25 round to the nearest hundered

jeyben [28]

34.30. I think this is the answer.

4 0

4 years ago

Read 2 more answers

Simplify: <br>{(12)^1 + (13)^-1}/[(1/5)^-2 × {(1/5)^-1 + (1/8)^-1}^-1]

CaHeK987 [17]

Step-by-step explanation:

$\underline{\underline{\sf{➤\:\:Solution}}}$

$\sf \dashrightarrow \: \dfrac{ \left(\left(12 \right)^{ - 1} + \left(13 \right)^{ - 1} \right) }{\left( \dfrac{1}{5}\right) ^{ - 2} \times\left( \left( \dfrac{1}{5} \right) ^{ - 1} +\left( \dfrac{1}{8} \right) ^{ - 1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{1}{12} + \dfrac{1}{13} \right) }{\left( \dfrac{5}{1}\right) ^{ 2} \times\left( \dfrac{5}{1} + \dfrac{8}{1} \right) ^{ - 1}}$

LCM of 12 and 13 is 156

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{1 \times 13 = 13}{12 \times 13 = 156} + \dfrac{1 \times 12 = 12}{13 \times 12 = 156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{5 + 8}{1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{13}{156} + \dfrac{12}{156} \right) }{\ \dfrac{25}{1} \times\left( \dfrac{13}{1} \right) ^{ - 1}}$

$\sf \dashrightarrow \: \dfrac{ \left(\dfrac{13 + 12}{156} \right) }{\ \dfrac{25}{1} \times\dfrac{1}{13} }$

$\sf \dashrightarrow \: \dfrac{25}{156} \div \dfrac{25}{13}$

$\sf \dashrightarrow \: \dfrac{ \cancel{25}}{156} \times \dfrac{13}{ \cancel{25} }$

$\sf \dashrightarrow \: \dfrac{13}{156}$

$\sf \dashrightarrow \: \dfrac{1}{12}$

$\sf \dashrightarrow \: Answer = \underline{\boxed{ \sf{ \dfrac{1}{12} }}}$

━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\underline{\sf{★\:\:Laws\:of\: Exponents :}}}$

$\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}$

$\sf 2^{nd} \: Law =$

$\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}$

$\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}$

$\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}$

$\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}$

$\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1$

3 0

3 years ago

Read 2 more answers

I need shown work with numbers or an explanation if need.

sveticcg [70]

Let us denote the semi arcs as congruent angles. This means that angles FEJ and EFJ are congruent (That is, they have the same measure). Since angles FEJ and EFJ have the same measure, this implies that sides EJ and FJ are equal. Since angles EJK and FJH are supplementary angles to angle EJF, this implies that EJK and FJH have the same measure.

Using the Angle Side Angle (SAS) criteria, we determine that triangles EKJ and triangle FJH are congruent. This implies that sides EK and FH are equal and that angles EKJ and FHJ are congruent. Note that angle EKJ is the same as EKF and that FHJ is the same as FHE.

Once again, since angles EKF and FHJ are congruent, and angle EKD is supplementary to the angle EKJ when angle FHG is supplementary to angle FHJ, then we have that angles EKD and angle FHG are congruent.

Using again the SAS criteria, we determine that triangles EKD and FHG are congruent.

From this reasoning, we have proved the following facts:

Triangle DEK is congruent to triangl GFH

Angle EKF is congruent to angle FHE

Segment EK is the same as segment FH

4 0

1 year ago

A salesperson set a goal to earn $3,000 in June. He receives a base salary of $1,500 per month as well as a 10% commission for a

Levart [38]

18,000 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1

5 0

3 years ago