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2 years ago

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Emery's blueberry farm can produce 535 bushels of blueberries in 2 days. At this rate, how many bushels of blueberries can Emery

's farm produce in 9 days? A. 2,675 B. 1,872.5 C. 2,140 D. 2,407.5

2 answers:

nevsk [136]2 years ago

7 0

Answer: THE ANSWER IS D 2,407.5

Step-by-step explanation:

Nimfa-mama [501]2 years ago

6 0

It’s 2,407.5 I divided 535÷2 then multiplied it by 9

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Distributive property of <br>4x 567

Maksim231197 [3]

567 = 500+60+7

4*567 = 4*(500+60*7)

4*567 = 4*500 + 4*60 + 4*7 ... see note below

4*567 = 2000 + 240 + 28

4*567 = 2268

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note: multiply the outer term 4 by each term inside the parenthesis to use the distributive property. The general distributive property is a*(b+c) = a*b+a*c. This can be extended to a*(b+c+d) = a*b+a*c+a*d. You can have as many terms as you like inside the parenthesis.

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3 years ago

a rectangular lawn has an area of a^3 - 125. use the difference of cubes to find out the dimensions of the rectangle.

ANEK [815]

The area of a rectangle is the product of its dimensions

The dimensions of the rectangle are: $\mathbf{Length = a -5}$ and $\mathbf{Width = a^2 + 5a + 25}$

The area is given as:

$\mathbf{Area = a^3 - 125}$

Express 125 as 5^3

$\mathbf{Area = a^3 - 5^3}$

Apply difference of cubes

$\mathbf{Area = (a - 5)(a^2 + 5a + 5^2)}$

$\mathbf{Area = (a - 5)(a^2 + 5a + 25)}$

The area of a rectangle is:

$\mathbf{Area = Length \times Width}$

So, by comparison:

$\mathbf{Length = a -5}$

$\mathbf{Width = a^2 + 5a + 25}$

Read more about areas at:

brainly.com/question/3518080

6 0

2 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$

6 0

3 years ago

AB and CD with M as the midpoint of both AB and CD AB = 6.4 cm and CD = 4.0 cm , A,B and C are not collinear…. Help !

loris [4]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

Apparently, you just need to draw a picture. This works well if you start with graph paper that is graduated in millimeters.

Of course, with a compass and ruler, you don't need graph paper. The lines don't need to be at right angles, either.

4 0

2 years ago

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were

zloy xaker [14]

Answer:

The required probability is 0.927

Step-by-step explanation:

Consider the provided information.

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.

That means 95% of students didn't enrolled in SAT prep course.

Let P(SAT) represents the enrolled in SAT prep course.

P(SAT)=0.05 and P(not SAT) = 0.95

30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.

P(F) represents the first choice college.

The probability he didn't take an SAT prep course is:

$P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}$

Substitute the respective values.

$P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}$

$P[\text{not SAT} |P(F)]\approx0.927$

Hence, the required probability is 0.927

6 0

2 years ago