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Marta_Voda [28]

3 years ago

7

Mark's soup recipe makes 6 servings and uses 4 carrots and 2 potaoes. How many potaoes and carrots does mark need for 15 serving

s?

2 answers:

lys-0071 [83]3 years ago

7 0

He would need 10 carrots and 5 potatoes for 15 servings.

Hope this helps.

finlep [7]3 years ago

7 0

Answer:

10 carrots and 5 potatoes

Step-by-step explanation:

<em>1) </em>15÷6=2.5

<em>2)</em>4 x 2.5=10

<em>3)</em>2 x 2.5=5

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Navigate three times the difference of 12 and twice some number is equal to 36. find the number.

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3 years ago

A parallelogram is shown below:

lana66690 [7]

Answer:

a). Area of the parallelogram = 8 cm²

b). Area of both the triangles = 4 cm²

Step-by-step explanation:

a). Area of the given parallelogram ABCD = Base × Height

= DC × (Vertical distance between

AB and DC)

= 9 × $\frac{8}{9}$

= 8 ft²

b). If we decompose this parallelogram into two triangles ΔABC and ADC by a diagonal AC,

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Therefore, area of ΔABC = Area of ΔADC = 4 ft²

7 0

2 years ago

Read 2 more answers

Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

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

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

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

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

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

so that the integral we want to find can be estimated as

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

It turns out that

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

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

5 0

3 years ago