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Anvisha [2.4K]
3 years ago
13

A hospital uses 218 liters of blood in a week.

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
5 0
B multiply 218 by 40 and you get 8,720
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Convert 467.03 to scientific notation
nikitadnepr [17]

Answer:

4.6703 x 10^2

3 0
3 years ago
Determine which point is the solution to the given system.
KatRina [158]

Answer:

D

Step-by-step explanation:

We equate both the y's and solve for x algebraically:

-\frac{4}{7}x+16=-\frac{2}{7}x+12\\16-12=-\frac{2}{7}x+\frac{4}{7}x\\4=\frac{2}{7}x\\x=\frac{4}{\frac{2}{7}}\\x=14

Now taking the value x= 14 and putting it in the 1st equation would give us y:

y=-\frac{4}{7}x+16\\y=-\frac{4}{7}(14)+16\\y=8

Hence the solution is (14,8), answer is D

7 0
4 years ago
Playland park charges 7 dollars admission plus 75 crnts per ride. Funland park charges 12.50$ admission plus 50 cents per ride.
WINSTONCH [101]

Answer:

22 rides

Step-by-step explanation:

Both the parks has a fixed cost (admission fee) and a variable cost (per ride cost). We can model 2 equations in "x" [let x be number of rides], equate them and find "x".

<u>Playland Park:</u>

Fixed Cost = 7

Variable Cost = 0.75x (in dollars)

Equation = 7 + 0.75x

<u>Funland Park:</u>

Fixed Cost = 12.50

Variable Cost = 0.50x (in dollars)

Equation = 12.50 + 0.50x

Now we equate and solve for x:

7+0.75x=12.50 + 0.50x\\0.75x-0.50x=12.50-7\\0.25x=5.50\\x=\frac{5.50}{0.25}\\x=22

Hence,

the cost would be same for both parks for 22 rides

8 0
3 years ago
What is the length of AC ?<br><br> 3ft<br> 4ft<br> 9ft<br> 18ft
WINSTONCH [101]

ΔACB and ΔMNB are similar. Therefore the corresponding sides are in proportion:

\dfrac{AC}{MN}=\dfrac{CB}{NB}\\\\MN=9\ ft\\CB=2\cdot3\ ft=6\ ft\\NB=3\ ft

Substitute:

\dfrac{AC}{9}=\dfrac{6}{3}\\\\\dfrac{AC}{9}=2\ \ \ \ |\cdot9\\\\AC=18\ ft

5 0
3 years ago
Plzzz I need help with this question I tried to solve it many times but I can't
blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as:  A1=\frac{(base)(height)}{2}

Area of a rectangle is calculated as: A2=(side)(side)

Area of a right trapezoid is: A3=\frac{(a+b)h}{2}, where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

S1=\frac{(2x-3)(4x-6)}{2}

S1=4x^{2}-12x+9

Area of rectangle S2:

S2 = (4x-6)(3x-2)

S2=12x^{2}-26x+12

Area of trapezoid S3:

S3=\frac{(2x+3+4x+1)(2x-3)}{2}

S3=\frac{(6x+4)(2x-3)}{2}

S3=6x^{2}-5x-6

2) a) S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)

S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6

S=10x^{2}-33x+37

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}

\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}

32x^{2}-48x-28x+42=12x^{2}+8x-18x-12

20x^{2}-66x+54=0

Using Bhaskara to solve the second degree equation:

\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}

x_{1}=\frac{66+6}{40} = 1.8

x_{2}=\frac{66-6}{40} = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) <u>Expand</u> <u>a</u> <u>polynomial</u> (or equation) is to multiply all the terms, remiving the parenthesis. <u>Reduce</u> <u>a</u> <u>polynomial</u> (or equation) is to combine terms alike,e.g.:

S=(2x-3)(5x-9)

S=10x^{2}-18x-15x+27 (expand)

S=10x^{2}-33x+27 (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

32x^{2}-48x-28x+42=12x^{2}+8x-18x-12+27

32x^{2}-48x-28x+42=12x^{2}+8x-18x+15

20x^{2}-66x+27=0

Solving:

\frac{66+\sqrt{(-66)^{2}-(4.20.27)} }{2(20)}

\frac{66+46.86}{40}

x_{1}=\frac{66+46.86}{40}= 2.82

x_{2}=\frac{66-46.86}{40} = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82

6 0
3 years ago
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