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3 years ago

I need a specific explanation of how to answer this

Mathematics

1 answer:

zimovet [89]3 years ago

5 0

A rectangle has two dimensions, width and length, and the area of it is their product.

since we know its area is 3x²-11x-4, then the two factors from that trinomial are the likely width and length, namely (3x + 1) (x - 4).

you can check them with FOIL.

You might be interested in

50 x 1 ten<br><br> What is the answer

My name is Ann [436]

Answer:

if its 50 X 10 then 500 if it is 50 X 1 then 50

Step-by-step explanation:

7 0

2 years ago

Solve for x by finding the missing side of the triangle. Round your answer to the nearest tenth.

MAXImum [283]

Answer:

Step-by-step explanation:

Tangent θ = Opposite / Adjacent

tan(29) = 20/x

Cross multiply.

x = 20 / (tan(29))

x = 22.544284

4 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

Savannah filled

denpristay [2]

Answear:

G=T hope dissssssssss helps

5 0

3 years ago

And here is the other one.

LUCKY_DIMON [66]

Since BD bisects angle ABC, that means angle ABD and angle CBD are equal to each other. With that set up the equation to solve for x like this:

-4x+33 = 2x+81
-2x -2x
————————
-6x +33 = 81
-33 -33
————————
-6x = 48
————- (divide by -6)
-6

x = -8

Now substitute that to ABD
-4(-8) +33
32 +33
=65

here’s CBD
2(-8) + 81
-16 + 81
=65

Finally angle ABC will be double the amount of ABD or CBD so 65 times 2 is 130.

ANSWERS: (angles)
ABD and CBD: 65
ABC: 130

3 0

3 years ago