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3 years ago

What is partial products

Mathematics

1 answer:

Irina-Kira [14]3 years ago

7 0

<span>the product of one term of a multiplicand and one term of its multiplier.</span>

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Kevin hired a moving company. The company charged $400 for its services, and Kevin gives the movers a 15% tip. How much does Kev

harina [27]

15% = 0.15

0.15 * 400 = 60

$60

7 0

2 years ago

John volunteers at the animal shelter on weekends. One Saturday, John unloaded 27 bags of dog food. Each bag weighed 35 pounds.

romanna [79]

Answer:

945 ounces

Step-by-step explanation:

if you read it closley you will find that if you multiply you will get 945 ounces because all you have to do is multipply.

5 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago

Yuri thinks that 3/4 is a root of the following function. q(x) = 6x3 + 19x2 – 15x – 28 Explain to Yuri why 3/4 cannot be a root.

k0ka [10]

Answer:

Yuri is not correct.

Step-by-step explanation:

Given expression is q(x) = 6x³ + 19x² - 15x - 28

If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0

Similarly, for x = $\frac{3}{4}$ ,

$q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28$

= $6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28$

= $(\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28$

= $(\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}$

= $-\frac{1666}{64}$

= $-\frac{833}{32}$ ≠ 0

Therefore, Yuri is not correct. x = $\frac{3}{4}$ can not be a root of the given expression.

6 0

3 years ago

Read 2 more answers

Find the volume of the triangular prism.

krok68 [10]

Approximately 673.76

4 0

3 years ago