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3 years ago

-4(x+9) - x > 3x - 14

Mathematics

1 answer:

stira [4]3 years ago

3 0

Answer:

x=9

Step-by-step explanation:

just trust me it is

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Number is 22 round down to

alina1380 [7]

Answer:

22 rounds down to 20 if that is what you are asking

Step-by-step explanation:

You did not finish your question i dont think. 22 DOES round down to 20 though.

3 0

3 years ago

2x - 5y=-15<br> can someone walk me through this?

HACTEHA [7]

The equation given in the question has two unknown variables in the form of "x" and "y". The exact value of "x" and "y" cannot be determined as two equations are needed to get to the exact values of "x" and "y". This equation can definitely be used to show the way for determining the values of "x" in terms of "y"and the value of "y" in terms of "x". Now let us check the equation given.
2x - 5y = - 15
2x = 5y - 15
2x = 5(y - 3)
x = [5(y - 3)]/2
Similarly the way the value of y can be determined in terms of "x" can also be shown.
2x - 5y = - 15
-5y = - 2x - 15
-5y = -(2x + 15)
5y = 2x + 15
y = (2x +15)/5
= (2x/5) + (15/5)
= (2x/5) + 3
So the final value of x is [5(y -3)]/2 and the value of y is (2x/5) + 3.

3 0

3 years ago

When Akiko measured a rose, its height was 5.8 in. After 10 weeks, the height was 1 1/3 times the original height. What was the

kramer

The solution is in the attachment

8 0

3 years ago

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

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

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago

A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?

adoni [48]

Answer:

$area \: = side \times side \\ area = 20 \times 20 \\ area = 400 \: {cm}^{2}$

8 0

3 years ago

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