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Shkiper50 [21]
3 years ago
11

Choose the monetary amount that is equivalent to 2 . $2.03 $0.30 $ 0.03 $ 2.30

Mathematics
1 answer:
kykrilka [37]3 years ago
5 0
$2.03 is the closest for being equivalent to 2
You might be interested in
PLEASE HELP
statuscvo [17]

Answer:

12%

Step-by-step explanation:

1 - 0.88= 0.12 x 100 = 12

7 0
3 years ago
For every 10 students there needs to be 1 chaperone during the trip to the 5 museum. Which of the following ratios does not repr
kipiarov [429]
1:10 does not represent the students to chaperone ratio because for every 10 students, there must be one chaperone.
6 0
3 years ago
Can someone help me with this question please!!!!!
faltersainse [42]
From the equation, it looks like we can set up a triangle like this.  One side is x and the other is x + 20 because the northbound car traveled 20 miles farther.  The hypotenuse is 100 because they were 100 miles apart.

Next, we can use the Pythagorean Theorem to solve for x:

a² + b² = c²
x² + (x + 20)² = 100²

This simplifies to:

x² + x² + 40x + 400 = 10000
2x² + 40x = 9600
2x² + 40x - 9600 = 0

After that, we can divided each side by 2:

x² + 20x - 4800 = 0

Next, we have to factor:

(x - 60)(x + 80) = 0
x = 60, -80

Since we know that distance can't be negative, 60 is the valid answer in this case.

x = 60
x + 20 = 80

Using this information, we know that the westbound car traveled 60 miles and the northbound car traveled 80 miles.

6 0
3 years ago
BRAINLIEST AND 20 POINTS ANSWER ASAP PLZ
galben [10]
R = m - v + 2, where r = faces, v = vertices, and m = edges

r = 28 - 13 + 2
r = 15 + 2
r = 17, so the first answer is correct.

7. The surface area of a cone is A = pi*r*sqrt(r^2 + H^2)

A = pi*(7)(sqrt(49 + 1849)
A = pi*(7)(43.57)
A = pi*305 = 959 m^2, so the first answer is correct.

13. The volume of the slab is V = HLW
V = (5 yards)(5 yards)(1/12 yards)
V = 25/12 cubic yards

So it costs $46.00*(25/12) = $95.83 of total concrete. The third answer is correct.

21. First, find the volume of the rectangular prism. V = HLW
V = (15 cm)(5 cm)(7 cm)
V = 525 cm^3

Next, find the volume of the pyramid. V = 1/3(BH), where H is the height of the pyramid and B is the area of the base of the pyramid. Note that B = (15 cm)(5 cm) = 75 cm^2

V = (1/3)(75 cm^2)(13 cm)
V = 325 cm^3

Add the two volumes together, the total volume is 850 cm^3. The fourth answer is correct.

22. The volume of a square pyramid is V = 1/3(S^2)(H), where S is the side length and H is the height.

V = (1/3)(20^2 in^2)(21 in)
V = 2800 in^3

Now that we know the volume of this pyramid, and that both pyramids have equal volume, we plugin our V to the equation for the volume.

2800 = (1/3)(84)(S^2)
2800 = 28S^2
100 = S^2
<span> 10 in = S, so we have a side length of 10 in, and the first answer is correct. </span>
3 0
3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

4 0
3 years ago
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