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4 years ago

(2x - 3y)(4x - y) can you help me

Mathematics

2 answers:

professor190 [17]4 years ago

5 0

You must FOIL (first, outside, inside, last)

First (The first number/variable of each parentheses must be multiplied together)

(2x - 3y)(4x - y)

$8x^{2}$

Outside (The outside number/variable of each parentheses must be multiplied together)

(2x - 3y)(4x - y)

-2xy

Inside (The inside number/variable of each parentheses must be multiplied together)

(2x - 3y)(4x - y)

-12xy

Last (The last number/variable of each parentheses must be multiplied together)

(2x - 3y)(4x - y)

$3y^{2}$

Now add/subtract each FOIL answer according to its sign

$8x^{2}$ - 2xy - 12xy + $3y^{2}$

Combine like terms

$8x^{2}$ + $3y^{2}$ - 14xy

Hope this helped!

~Just a girl in love with Shawn Mendes

Mnenie [13.5K]4 years ago

3 0

Answer:

Step-by-step explanation:

Use FOIL: <em>(a + b)(c + d) = ac + ad + bc + bd</em>

(2x - 3y)(4x - y) = (2x)(4x) + (2x)(-y) + (-3y)(4x) + (-3y)(-y)

= 8x² - 2xy - 12xy + 3y²

<em>combine like terms</em>

= 8x² + (-2xy - 12xy) + 3y² = 8x² - 14xy + 3y²

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A cow gave 8 1 2 liters of milk in the morning, 3 5 liters less than this at the noon milking, and 6 3 10 liters in the evening.

Nitella [24]

Answer:

$22\frac{7}{10}\ l$ or $22.7\ l$

Step-by-step explanation:

<u><em>The correct question is</em></u>

A cow gave 8 1\2 liters of milk in the morning, 3\5 liters less than this at the noon milking, and 6 3/10 liters in the evening. How many liters of milk did the cow give during the entire day?

we know that

The total liters of milk given by the cow during the entire day is equal to the amount in the morning plus the amount at the noon plus the amount in the evening.

Let

x -----> liters of milk given by the cow in the morning

y ----> liters of milk given by the cow at the noon

z -----> liters of milk given by the cow in the evening

T ----> total liters of milk given by the cow during the entire day

T=x+y+z

we have

$x=8\frac{1}{2}\ l=\frac{8*2+1}{2}=\frac{17}{2}=\frac{85}{10}\ l$

$y=\frac{17}{2}-\frac{3}{5}=\frac{85-6}{10}=\frac{79}{10}\ l$

$z=6\frac{3}{10}\ l=\frac{6*10+3}{10}=\frac{63}{10}\ l$

$T=\frac{85}{10}+\frac{79}{10}+\frac{63}{10}$

$T=\frac{227}{10}\ l$

$T=22.7\ l$

Convert to mixed number

$22.7\ l=22+0.7=22+\frac{7}{10}=22\frac{7}{10}\ l$

6 0

3 years ago

Find The Domain Find the Value

laila [671]

1a) The domain is a set of all real numbers not equal to 5

1b) The domain is a set of all real numbers not equal to 4

<h3>What is the domain of the function?</h3>

Domain of a function is the set of all possible input values that make the function defined.

1a) We are given the function;

C(x) = -3²/(x - 5)

The domain will be a set of values that makes the denominator not equal to zero. Thus;

The domain is a set of all real numbers not equal to 5

1b) The function is; R(x) = 4x/(3x - 12)

The domain will be a set of values that makes the denominator not equal to zero. Thus;

The domain is a set of all real numbers not equal to 4

2a) Function is;

f(x) = (x + 8)/(2x - 1)

f(2) = (2 + 8)/(2(2) - 1)

f(2) = 10/3

f(0) = (0 + 8)/(2(0) - 1)

f(0) = -8

f(-1/2) = (-1/2 + 8)/(2(-1/2) - 1)

f(-1/2) = -15/4

2b) Function is;

r(t) = (t² + 8)/(t³ - 25t)

r(3) = (3² + 8)/(3³ - 25(3)) = -17/48

r(2) = (2² + 8)/(2³ - 25(2)) = -12/42 = -2/7

r(-1) = ((-1)² + 8)/((-1)³ - 25(-1)) = 9/24 = 3/8

Read more about Domain of function at; brainly.com/question/10197594

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3 0

1 year ago

During the lunar cycle, the Moon always appears: A. to wax, but not wane. B. to wane, but not wax. C. to wax, then wane. D. to w

JulijaS [17]

Answer:

C.to wax then wane

Step-by-step explanation:

Waxing is "Maxing," or growing with light until a full moon is reached. Waning is the opposite, or decreasing after a full moon, and is always illuminated on the left. Then, there is a waxing or waning Gibbous moon, which means more than half of the moon is illuminated.

5 0

3 years ago

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

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