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3 years ago

To solve the equation x + 7 =15 you would need to......

Mathematics

1 answer:

zysi [14]3 years ago

4 0

Subtraction property of equality. Subtract both sides by 7, giving you the answer of 8.

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PLEASE TELL ME HOW YOU GOT THE ANSWER!!!!!

Lesechka [4]

C is the answer. if the water is being pumped out then the height of the water would decrease.

8 0

3 years ago

HURRY PLEASE HELP ME? WiLL GIvE brainliest.

solniwko [45]

Subtitute the pair (3,28) to x and y, if the result in right side is -8, then the pair is a solution to the equation

2x - y/2
= 2(3) - 28/2
= 6 - 14
= -8

The result is -8, so the pair (3,28) is a solution

3 0

3 years ago

Determine the equation of the line that has an x-intercept of 6 and passes through the point (6,4)

butalik [34]

Putting a = 6 and b= 3/2 in eqn. (1) the required equation is:-

x /6 + 2.y/3 = 1. or, x + 4y = 6. Answer.

5 0

2 years ago

Read 2 more answers

A fruit seller purchased 20 kg of apples at Rs 80 per kg and sold them at Rs 85 per kg. find profit and profit percent

viktelen [127]

Answer:

Profit = Rs 100

Profit %age = 6.25%

Step-by-step explanation:

Cost Price = Rs 80

Cost Price for 20 kg = Rs 1600

Selling Price = Rs 85

Profit for 1 kg = Rs 85 - Rs 80

=> Rs 5 (For 1 Kg)

Profit for 20 kg:

=> Rs 5*20

=> Rs 100

Now Profit %age:

=> $\frac{100}{1600}* 100$

=> 0.0625 * 100

=> 6.25%

7 0

3 years ago

Read 2 more answers

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

3 years ago