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vlabodo [156]
3 years ago
12

What is 3+3×3-3+3 equal?

Mathematics
2 answers:
tangare [24]3 years ago
5 0
Use Pemdas it would be much easier to do that
LiRa [457]3 years ago
4 0
12
Remember to use PEMDAS when solving these problems. You multiply before you work with addition and subtraction.
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Write this as an equation, then solve.
ANTONII [103]
19.5 in difference 
difference indicates that you'll be subtracting so 29 3/4 - 10 1/4 = 19.5

3 0
3 years ago
The hypotenuse AB of the right triangle ABC is parallel to the axis of the abscess. Find the length of the hypotenuse if A (-1;
lukranit [14]

Answer:

  10.25

Step-by-step explanation:

The slope of AC is given by the slope formula:

  m = (y2 -y1)/(x2 -x1)

  m = (-4 -1)/(3 -(-1)) = -5/4

Then the slope of CB is the opposite reciprocal, 4/5. The equation of line CB in point-slope form is ...

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

  y -(-4) = 4/5(x -3) . . . . line CB

When y = 1 (to match the y-value of A), then ...

  1 +4 = 4/5(x -3)

  5(5/4) = (x -3) . . . . . multiply by 5/4

  6.25 +3 = x = 9.25 . . . . add 3

Point B is (9.25, 1).

The length of the hypotenuse is ...

  9.25 -(-1) = 10.25

6 0
2 years ago
NEED HELP! Thanks!
sladkih [1.3K]
I think the answer would be 7.5 hours or 4 hours,not so sure which one exactly

6 0
3 years ago
What is a histogram?
aliya0001 [1]

Answer:

a diagram consisting of rectangles whose area is proportional to the frequency of a variable and whose width is equal to the class interval.

explanation: A histogram is a chart that shows frequencies for. intervals of values of a metric variable. Such intervals as known as “bins” and they all have the same widths. The example above uses $25 as its bin width. So it shows how many people make between $800 and $825, $825 and $850 and so on.

5 0
3 years ago
Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
4 years ago
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