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Alex Ar [27]
3 years ago
12

Select all of the possibilities for the number of solutions of a two variable linear system

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
5 0

Select all of the possibilities for the number of solutions of a two-variable linear system.

YES no solutions

YES 1 solution

NO 2 solutions

YES infinite solutions

NNADVOKAT [17]3 years ago
4 0

Answer:

1, 2 and 4

if parrallel lines: no solution

Step-by-step explanation:

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VMariaS [17]
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3 0
3 years ago
Predict: How do you think the formula for area of a triangle will relate to the formula for area of a rectangle? Do you
densk [106]

Answer:

I think is the same formula area .A triangle is half as big as the rectangle that surrounds it, which is why the area of a triangle is one-half base times height.

Step-by-step explanation:

The area of a rectangle is given by the relation length* width. The area of a triangle is given by the relation (1/2)*base* height. Now we know that the two have the same area equal to 150 cm^2. Also, when placed on the same level the tip of the triangle touches the upper edge of the rectangle.

3 0
3 years ago
Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

4 0
3 years ago
Montoya Construction needs to borrow $375,000 to build a road to install utilities in a small subdivision. It borrows the funds
MissTica

Step 1

Given;

\begin{gathered} \text{Principal(p)= \$375000} \\ \text{First rate = 8\%=}\frac{8}{100}=0.08 \\ \text{Second rate = 20\%= }\frac{20}{100}=0.2 \\ \text{Time}=\frac{90}{365}=\frac{18}{73} \end{gathered}

Required; To find the difference in interest between the two periods.

Step 2

State the formula for simple interest

A=P(1+rt)

Step 3

Find the interest when the rate is 8%

\begin{gathered} A=375000(1+(0.08\times\frac{18}{73}) \\ A=375000(1+\frac{36}{1825}) \\ A=\text{\$}382397.26 \end{gathered}

Therefore the interest is given as;

A-P=382397.26-375000=\text{\$}7397.26

Step 4

Find the interest in 1980 with a 20% rate

\begin{gathered} A=375000(1+(0.2\times\frac{18}{73}) \\ A=\text{\$}393493.15 \end{gathered}

The interest is given as;

A-p=393493.15-375000=\text{\$}18493.15\text{ }

Step 5

Find the difference in interest between the two rates.

\text{\$}18493.15-\text{\$}7397.26=\text{\$}11095.89

Hence, the difference in interest between the two rates = $11095.89

4 0
1 year ago
Find the solution of this system of equations 5x - 6y =-90 -10x - 6y =-90
ExtremeBDS [4]

5x - 6y = -90

-10 x - 6y = -90

Subtracting,

15x = 0

x = 0

y = -90/(-6) = 15

Answer: x=0, y=15


8 0
3 years ago
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