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4 years ago

How would i calculate 1.9 minus 11/12?

1 answer:

8 0

Step 1) turn 1.9 into an improper fraction.

= 19/10

step 2) find a least common denominator between 10 and 12.

= 60
(10x6 and 12x5)

step 3) convert the fractions so they have the same denominator.

= 19x6 is 114.
10x6= 60.
= 11x5 is 55.
12x5 is 60.
now you have your new fractions.

step 4) subtract.

= 114/60 - 55/60 is 59/60.

step 5) simplify.

the answer is either 0.9833... (3 repeating) or just 59/60 (you can't simplify the fraction any further).

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Fill the missing values to make the equations true.

taurus [48]

Logarithm properties:

$\displaystyle \log_a b+\log_a c=\log_a (b\cdot c)\\\\\log_ab-\log_a c=\log_a \left( \frac{b}{c} \right)\\\\\log_{a^p}b^q= \frac{q}{p} \log_a b$

According to this, we can get:

$\displaystyle a)\log_7 3+\log_7 8=\log_7 (3\cdot 8)=\log_724\\\\b) \log_89-\log_85=\log_8 \frac{9}{5}\\\\c)\log_325=2\log_35$

3 0

3 years ago

Helllppppppppppppppppppppppppppppp

Georgia [21]

The equation: n - 4 = -15

Now let's solve our equation for n.

n - 4 = -15

Add 4 to both sides.

n = -11

4 0

4 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

Tickets for a basketball game cost $2 for students and $5 for adults. The number of students was 3 less than 10 times the number

34kurt

2x+5y=619
x=10y-3
________
2(10y-3) +5y=619
20y-6+5y=619
25y=619+6
25y=625
y=625/25
y=25
x=10*25-3
x=250-3
x=247

6 0

3 years ago

Read 2 more answers

Point Q'Q ′ Q, prime is the image of Q(0,6)Q(0,6)Q, left parenthesis, 0, comma, 6, right parenthesis under the translation (x,y)

Marina86 [1]

Q is located at (0,6)

The translation rule is $(x,y) \to (x+7,y-5)$ which says to add 7 to the x coordinate and subtract 5 from the y coordinate. Doing that to (0,6) moves it to (7,1) which is where point Q' is located.

In other words, if you shift point Q(0,6) seven units to the right and five units down, then it arrives at Q ' (7,1)

<h3>Answer: (7, 1)</h3>

8 0

3 years ago