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2 years ago

-4 = p + 7 using sum in verbal expression form

Mathematics

2 answers:

vlada-n [284]2 years ago

8 0

P=-11. Explanation: -4=p+7
-7. -7
-11=p

hjlf2 years ago

3 0

Negative four equals the sum of p and seven

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What is the area of circle with a radius of 13 milimeters? use 3.14 for pi

MrRissso [65]

Answer:

530.66 mm²

5.3066 cm²

Step-by-step explanation:

A = π·r²

= 3.14×(13mm)²

= 3.14×169mm²

= 530.66 mm²

= 5.3066 cm²

8 0

3 years ago

Read 2 more answers

The bike store marks up the wholesale cost of all of the bikes they sell by 30%.

aev [14]

Answer:

A ) The money which Andre pay for the bike is $ 67.30

B ) The money which Andre pay for the bike is $ 76.92

Step-by-step explanation:

Given as :

The wholesale cost or market price of the bike = m.p = $ 96.15

A ) The discount percentage = 30 %

Let The selling price of the bike after discount = s.p

Now ,

Discount % = $\dfrac{m.p - s. p }{m . p}$

or, 30 % = 1 - $\dfrac{s. p }{m . p}$

or, $\dfrac{30 }{100}$ = 1 - $\dfrac{s. p }{96.15}$

Or, $\dfrac{s. p }{96.15}$ = 1 - $\dfrac{30 }{100}$

or, $\dfrac{s. p }{96.15}$ = $\dfrac{100 - 30 }{100}$

Or, $\dfrac{s. p }{96.15}$ = $\dfrac{70}{100}$

Or, s. p = $\dfrac{70}{100}$ × 96.15

∴ s.p = $ 67.305

So, He will pay $ 67.30

Hence The money which Andre pay for the bike is $ 67.30 . Answer

B ) If the bike is discounted by 20 % , let The selling price = s.p'

So,

Discount % = $\dfrac{m.p - s. p }{m . p}$

or, 20 % = 1 - $\dfrac{s. p' }{m . p}$

or, $\dfrac{20 }{100}$ = 1 - $\dfrac{s. p' }{96.15}$

Or, $\dfrac{s. p' }{96.15}$ = 1 - $\dfrac{20 }{100}$

or, $\dfrac{s. p' }{96.15}$ = $\dfrac{100 - 20 }{100}$

Or, $\dfrac{s. p' }{96.15}$ = $\dfrac{80}{100}$

Or, s. p' = $\dfrac{80}{100}$ × 96.15

∴ s.p' = $ 76.92

So, He will pay $ 76.92

Hence The money which Andre pay for the bike is $ 76.92 . Answer

7 0

3 years ago

Solve this query plzzz

Olin [163]

Step-by-step explanation:

$\frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} = \tan \:A \\ \\ LHS = \frac{\sin \:4A}{\cos \:2A} \times \frac{1 - \cos \:2A}{1 - \cos\:4A} \\ \\ = \frac{2\sin \:2A.\cos \:2A}{\cos \:2A} \times \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\ \\ = 2\sin \:2A \times \frac{1 - 2 { \cos}^{2}A + 1}{1 - 2 { \cos}^{2}2A + 1 } \\ \\ = 2\sin \:2A \times \frac{2- 2 { \cos}^{2}A }{2 - 2 { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{2(1 - { \cos}^{2}A) }{2 (1- { \cos}^{2}2A) } \\ \\ = 2\sin \:2A \times \frac{1 - { \cos}^{2}A}{1- { \cos}^{2}2A } \\ \\ = 2\sin \:2A \times \frac{ { \sin}^{2}A}{{ \sin}^{2}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ \sin}2A } \\ \\ = 2 \times \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \: A } \\ \\ = \frac{ { \sin}A}{ \cos \: A } \\ \\ = tan \: A \\ \\ = RHS \\$