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defon
4 years ago
10

Olve. Do your work on a separate she

Mathematics
2 answers:
Mazyrski [523]4 years ago
5 0

Answer:

Step-by-step explanation:

2 coins are 0.25 = 50 cents. That leaves 92-50 = 42 cents and 5 coins to go.

The only way to get 2 cents is to have 2 pennies. That leaves 40 cents and 3 coins to go.

The only way to get 40 cents with 3 coins is 1 quarter, 1 dime, and 1 nickel.

So you have 3 quarters = 75 cents.

2 pennies = 2 cents.

1 dime = 10 cents.

1 nickel = 5 cents.

92 cents total and 7 coins.The only way to get 40 cents

fredd [130]4 years ago
4 0

1 nickel, 1 dime, 3 quarters, and 2 pennies is the correct answer. 0.05 + 0.10 + 0.75 + 0.02 = 0.92. Plus there are only 7 coins and at least 2 quarters.

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What is the value of 26+5-4*3
MatroZZZ [7]

You have to use PEMDAS (Parentheses Exponents Multiplication/Division Addition/Subtraction) to solve this,

Because you don't have any parenthesis or exponents you start with Multiplication/Division. This means you do 4*3 to get 26+5-12

Then, you move to Addition/Subtraction in which you complete left to right, So you do 26+5 to get 31-12

Finally you subtract 31-12 to get 19

8 0
4 years ago
A pitcher could hold 1/8 of a gallon of water. If John filled up 9 Pitchers, how much water would he have.
IgorLugansk [536]
1 pitcher = 1/8 gallon
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6 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
4. What is the slope of the line through the points (2, 5) and (6, 13)?
julsineya [31]

the slope of the line is 2

4 0
3 years ago
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Marizza181 [45]
I believe it is 12.56 sq. Units
7 0
4 years ago
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