The inequality of 8z+3-2z<51 is

A bicycle helmet has a selling price of $50. The helmet is discounted 30% off the selling price. The sales tax rate is 6%. What

Andrei [34K]

Convert 30% to a decimal by dividing by 100 = 0.30

Multiply by the cost of the helmet, $50
50 x 0.30 = $15, the amount of the discount.

50-15 = $35, the sale price of the helmet.

For the sales tax, again, convert to a decimal by dividing by 100 = 0.06

35 x 0.06 = 2.10, the amount of sales tax

$35 + $2.10 = $37.10, total cost with tax

6 0

2 years ago

there are 40 players in 2 basketball teams.25% of the players did not like thier positions .how many players like their position

scZoUnD [109]

Answer:

0.25 = 25%

40 players * 0.25 percent = 10 players

10 players did <u><em>NOT </em></u> like their positions.

(the 2 teams is a trick to throw you off)

4 0

3 years ago

Read 2 more answers

a rectangular prism has a volume of 1374.28 cubic inches.The prism is 9.4 inches wide and 17.2 inches long.What is the height of

NISA [10]

The volume V of a prism can be found by:
The volume of a rectangular prism can be found through:
$V=lwh$
Where l is the length, w is the width, and h is the height.

The height can be found by rearranging the equation to:
$h= \frac{V}{wl}$

Take the volume (1374.28) and divide it by the length times the width (9.4*17.2=161.68).
1374.28/161.68= 8.5 inches is the height of the prism.

6 0

3 years ago

Help please and thank you

aksik [14]

Remember
$(a^b)^c=a^{bc}$
and
$\frac{x^a}{x^b}=x^{a-b}$

so

$\frac{(7^2)^5}{7^{-6}}=$
$\frac{7^{2*5}}{7^{-6}}=$
$\frac{7^{10}}{7^{-6}}=$
$7^{10-(-6)}=$
$7^{10+6}=$
$7^{16}$

the answer is D

5 0

3 years ago

Prove that the diagonals of a parallelogram bisect each other

Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof : $\begin{array}{ |c| c | c | } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In \: \triangle ^{s} \:AOB \: and \: COD } \\ \sf{(i)}& \sf{ \angle \: OAB = \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &\sf{ \angle \: OBA= \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. Proved ✔$

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

$\underbrace{ \overbrace {\mathfrak{Carry \: On \: Learning}}}$ ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0

3 years ago