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Reika [66]
3 years ago
5

Malika charges $7.20 per hour for babysitting.she babysits 4 hours each week.she spends $10 per week and puts the rest of her ea

rnings in a saving account.after 12 weeks,Malika's grandparents put a gift of money in her savings account that is equal to 0.25 times the amount she has saved from babysitting earning.How much money does Malika have in her savings account after the gift from her grandparents? Show your work
Mathematics
1 answer:
DochEvi [55]3 years ago
4 0
78. Malika earns 7.20 dollars per hour for babysitting. She Baby Sit for 4 hours each week. Now, she had a total of 12 weeks of working => 7.20 * 4 hours = 28.8 dollars => 28.8 dollars * 12 = 345.6 dollars. She spent 10 dollars per week => 10 * 12 = 120 dollars => 345.6 – 120 = 225.6 dollars is the money that goes to her savings. Her grandparents put 0.25 times of her money she put on her savings. => 225.6 * .25 = 56.4 dollars => 225.6 + 56.4 = 282 dollars.
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VashaNatasha [74]

Answer:

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6 0
3 years ago
What is 18/25 as a decimal
Serga [27]

Answer: 0.72

Step-by-step explanation:

Hope that helps :)

4 0
3 years ago
Read 2 more answers
Can someone help last one :((
vovangra [49]

Answer:

40.49cm²

Step-by-step explanation:

I hope it helps

8 0
3 years ago
Rectangle R has varying length l and width w but a constant perimeter of 4 ft. A. Express the area A as a function of l. What do
ivanzaharov [21]
Given:
l = length of the rectangle
w = width of the rectangle
P = 4 ft, constant perimeter

Because the given perimeter is constant,
2(w + l) = 4
w + l = 2
w = 2 - l            (1)

Part A.
The area is
A = w*l 
   = (2 - l)*l
 A  = 2l - l²
This is a quadratic function or a parabola.

Part B.
Write the parabola in standard form.
A = -[l² - 2l]
   = -[ (l -1)² - 1]
   = -(l -1)² + 1
This is a parabola with vertex at (1, 1). Because the leading coefficient is negative the curve is downward, as shown below.

The maximum value occurs at the vertex, so the maximum value of A = 1.
From equation (1), obtain
w = 2 - l = 2 - 1 = 1.
The maximum value of the area occurs when w=1 and l=1 (a square).

Answer:
The area is maximum when l=1 and w=1.
The geometric argument is based on the vertex of the parabola denoting maximum area.

4 0
4 years ago
Solve 4 cos2x-3 = 0 for all real values of x.
vichka [17]

4 cos² x - 3 = 0

4 cos² x = 3

cos² x = 3/4

cos x = ±(√3)/2

Fixing the squared cosine doesn't discriminate among quadrants.  There's one in every quadrant

cos x = ± cos(π/6)

Let's do plus first.  In general, cos x = cos a has solutions x = ±a + 2πk integer k

cos x = cos(π/6)

x = ±π/6 + 2πk

Minus next.

cos x = -cos(π/6)

cos x = cos(π - π/6)

cos x = cos(5π/6)

x = ±5π/6 + 2πk

We'll write all our solutions as

x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k

3 0
3 years ago
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