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3 years ago

One-quarter of a number is 6 work out one-third of the number

Mathematics

1 answer:

expeople1 [14]3 years ago

6 0

Answer:

the number is 24. One third of the number is 8.

Step-by-step explanation:

Let the number be x.

(1/4) * x = 6

x = 6 * 4 = 24

x = 24

one third of the number :

(1/3) * 24 = 24/ 3 = 8

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In how many ways can the letters in the word spoon be arranged?

Kisachek [45]

I search and found different answers but the nearest is 24

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2 years ago

You are measuring the height of a statue. You stand 17 feet from the base of the statue. You measure the angle of elevation from

olga_2 [115]

Answer:

height ≈ 31 ft

Step-by-step explanation:

The illustration below will form a right angle triangle. He is standing 17 ft away from the base of the statue. The angle of elevation from the ground to the top of the statue is 61°. The height of the statue can be computed below.

adjacent side of the triangle = 17 ft

opposite side = a

Using SOHCAHTOA principle

tan 61° = opposite/adjacent

tan 61° = a/17

cross multiply

a = 17 tan 61°

a = 17 × 1.80404775527

a = 30.6688118396

a ≈ 31 ft

6 0

3 years ago

Use the Pythagorean Theorem to find the distance from Emma's house to the library.

AlladinOne [14]

Answer:square root 130

Step-by-step explanation:

4 0

3 years ago

25 POINTS!! HELP ASAP PLEASE!!!

gladu [14]

Answer:

Step-by-step explanation:

Stem is the first digit and the leaf is the last digit.

Below the table is explained that 1 l 7 is 1.7 because it also could mean 17. So with that clear the terms from the Stem l Leaf are as follow:

1 combined with 3, 7 and 7:

1.3, 1.7 and 1.7

2 combined with 6, 8 and 8:

2.6, 2.8, 2.8

3 combined with 5 and 7:

3.5 and 3.7

All the terms:

1.3, 1.7, 1.7, 2.6, 2.8, 2.8, 3.5, 3.7

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.

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