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3 years ago

How to simplify 2(3-4a)+5(a-7)

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

4 0

Multiply the first bracket by 2

Multiply the second bracket by 5

6-8a+5a-35

6-35-8a+5a ( just rearranging)

-29-3a or -3a-29 same answer just rearranging.

Answer is -29-3a or -3a-29

Aleks [24]3 years ago

3 0

Answer:

2(3-4a)+5(a-7)

=6-8a+5a-35

=-3a-29

Step-by-step explanation:

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Trucks are delivering gravel to a construction site each truck hold 7.5 cubic yards of gravel. The weight of 1 cubic yard of gra

Romashka-Z-Leto [24]

The first thing you must do is to calculate the weight of the gravel (w) that the truck holds (7.5 yards³). So, you have:

If the weight of 1 yard³ is 1.48 tons, then the weight of 7.5 yards³ is:

w=7.5x1.48 tons
w=11.1 tons

The problem says that the gravel will be placed in containers that each hold 3.7 tons of gravel,so, If they need 1 cotainer to place 3.7 tons, how many containers they need to place 11.1 tons?

1 container-------3.7 tons of gravel
x-------11.1 tons of gravel

x=(11.1x1)/3.7
x=11.1/3.7
x=3 containers

How many containers of this size are needed to hold all the gravel from one truck?

The answer is: 3 containers.

7 0

4 years ago

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

WITCHER [35]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

The picture of the question in the attached figure

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)\\C=6\pi\ cm$

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

$AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

4 0

4 years ago

A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a t

USPshnik [31]

Answer:

The probability is $P(K) = \frac{ 28 }{195}$

Step-by-step explanation:

From the question we are told that

The number of green marbles is $n_g = 3$

The number of red marbles is $n_b = 5$

The number of red marbles is $n_r = 7$

Generally the total number of marbles is mathematically represented as

$n_t = n_r + n_g + n_ b$

$n_t = 7 + 3 + 5$

$n_t = 5$

Generally total number of marbles that are not red is

$n_k = n_g + n_ b$

=> $n_k = 3 + 5$

=> $n_k = 8$

The probability of the first ball not being red is mathematically represented as

$P(r') = \frac{n_k}{n_t}$

=> $P(r') = \frac{ 8}{15}$

The probability of the second ball not being red is mathematically represented as

$P(r'') = \frac{n_k - 1}{n_t -1}$

=> $P(r'') = \frac{ 8 -1 }{15-1}$ (the subtraction is because the marbles where selected without replacement )

=> $P(r'') = \frac{ 7 }{14}$

The probability that the first two balls is not red is mathematically represented as

$P(R) = P(r') * P(r'')$

=> $P(R) = \frac{ 8}{15} * \frac{ 7 }{14}$

=> $P(R) = \frac{ 8 }{30}$

The probability of the third ball being red is mathematically represented as

$P(r) = \frac{n_r}{ n_t -2}$ (the subtraction is because the marbles where selected without replacement )

$P(r) = \frac{7}{ 15 -2}$

=> $P(r) = \frac{7}{ 13}$

Generally the probability of the first two marble not being red and the third marble being red is mathematically represented as

$P(K) = P(R) * P(r)$

$P(K) = \frac{ 8 }{30} * \frac{7}{ 13}$

=> $P(K) = \frac{ 28 }{195}$

6 0

3 years ago

Please help I’m timed

kupik [55]

It’s the second one :)

3 0

3 years ago

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The reflection of a Quadrant II point is located in Quadrant I. Assuming no error was made, what kind of reflection occurred?

gizmo_the_mogwai [7]

Answer:

A reflection across the y- axis

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y) → (-x, y)

A point in the second quadrant (- x, y)

Under reflection in the y- axis is (x, y) ← point in first quadrant

3 0

3 years ago

Read 2 more answers