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belka [17]
3 years ago
12

Chen uses different strategies to add. He works with the addends 4,5,6,7

Mathematics
1 answer:
Lelu [443]3 years ago
4 0
You can do 6 I guess. Then do 6+6=12
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Dado un triángulo rectángulo cuyo perímetro es de 48 cm, la diferencia entre su lado mayor y su lado menor es de 8 cm y su área
lesya692 [45]

Answer:

5471002836nmcncm.....

5 0
2 years ago
When multiplying numbers in scientific notation you blank the coefficients and substract the exponents?
34kurt
For the sake of example, let's multiply the two numbers 2.3 \times 10^5 and 3.5 \times 10^7 together. Altogether, we have:

2.3\times10^5\times3.5\times10^7

Rearranging the expression, we can group the exponents and coefficients together:

2.3\times3.5\times10^5\times10^7

Multiplying each out, we notice that since 10^5 and 10^7 have the same base (10), multiplying them has the effect of adding their exponents, which leaves us with:

2.3\times3.5\times10^{5+7}=8.05\times10^{12}

The takeaway here is that multiplying two numbers in scientific notation together has the effect of multiplying its coefficients and <em>adding</em> its exponents.
8 0
3 years ago
Please help me find the volume
SSSSS [86.1K]

Answer:

C,. 3 3/4

Step-by-step explanation:

length times with times hight

1 1/4 x 1 1/2 x 2

convert everything to same denominator

1 1/4

1 2/4

1 4/4

solve

1 1/4 x 1 2/4 x 1 4/4

4 0
3 years ago
A central angle of a circle measures 42 degrees. What else can be shown to measure 42 degrees?
Slav-nsk [51]

Answer:

Vertically opposite angles

Step-by-step explanation:

Given

\theta = 42^\circ

Required

What else can measure \theta = 42^\circ

The question has missing options. So, I will give a general explanation.

Using the attached image as a point of reference.

Given that:

\theta = 42^\circ

The value of b is calculated as:

b = \theta

Because b and \theta are vertically opposite angles

<em>This gives:</em>

<em></em>b = 42^\circ<em></em>

<em></em>

<em>So, one possible answer to the question is a vertically opposite angle</em>

4 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
3 years ago
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