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3 years ago

13

A gym membership is offered in January for 29.00 per month. If you pay upfront for the year you save 20% discount. How much mone

y do you save by paying upfront rather than monthly?

1 answer:

lukranit [14]3 years ago

8 0

Answer:

$69.60

Step-by-step explanation:

A gym membership is offered in January for $29.00 per month.

Monthly payment:

$\$29\cdot 12=\$348$

Upfront payment:

If you pay upfront for the year you save 20% discount, then you'll save 20% of $348 that is

$\$348\cdot 0.2=\$69.60$

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81

Step-by-step explanation:

3^4

=3*3*3*3

=81

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3 years ago

Solve for x 3x + 10 - 4 = 21

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Answer:

X = 5

Step-by-step explanation:

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3 years ago

Which of the following is not a

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B 7/8 shows division, not multiplication

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3 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

Help me please i’ll rate brainliest

Cloud [144]

Answer:

<em>- 1.568</em>

Step-by-step explanation:

Let's assume that the base of the given logarithm is 10.

log m = 0.345 ⇒ $10^{0.345}$ = m

log n = 1.223 ⇒ $10^{1.223}$ = n

mn = $10^{0.345}$ × $10^{1.223}$ = $10^{1.568}$

$\frac{1}{mn}$ = $10^{-1.568}$

log $10^{-1.568}$ = - 1.568 × log 10 = <em>- 1.568</em>

5 0

3 years ago