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Over [174]
3 years ago
6

Help with these 4 questions please and thanks!!!

Mathematics
1 answer:
HACTEHA [7]3 years ago
8 0
1. The radius is ✓33≈C
2. Center: (-2,3)
Radius: ✓23
3. Equation: (x+3)^2+(y+5)^2=125
4. Center: (-1,0)

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Which equation shows how the solution to 25x^2- 1=0 can be found by factoring?
stepladder [879]
D. Use FOIL and guess and check to factor the equation
5 0
3 years ago
Factorize x squared - 1x +10 =0
Natalka [10]

Answer:

no solution exists:

Step-by-step explanation:

x²−1x+10=0

Step 1: Simplify both sides of the equation.

x²−x+10=0

Step 2: Subtract 10 from both sides.

x²−x+10−10=0−10

x²−x=−10

Step 3: The coefficient of -x is -1. Let b=-1.

Then we need to add (b/2)^2=1/4 to both sides to complete the square.

Add 1/4 to both sides.

x²−x+ 1/4=−10+ 1/4

x²−x+ 1/4=−39 /4

Step 4: Factor left side.

(x  -1/2)² = −39 /4

Step 5: Take square root.

x  −1 /2 =±√ −39 /4

Step 6: Add 1/2 to both sides.

x  =−1/2  + 1/2 = 1/2 ±√ -39/4

x = 1/2 ±√ -39/4

No real solutions.

6 0
3 years ago
Need help please help
professor190 [17]
153.9 inches is the correct answer
5 0
3 years ago
Read 2 more answers
What is the value for Q1
Sever21 [200]

Answer:

its 5 and 1/2

Step-by-step explanation:

6 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.
Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]

The left and right endpoints of the i-th subinterval, respectively, are

\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8

r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8

for 1\le i\le4, and the respective midpoints are

m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8

  • Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}

  • Midpoint rule

We approximate the area for each subinterval by

M_i=f(m_i)(\ell_i-r_i)

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}

  • Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial p_i(x), where

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

so that

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It so happens that the integral of p_i(x) reduces nicely to the form you're probably more familiar with,

S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))

Then the integral is approximately

\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0
3 years ago
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