ANSWER THIS PLEASE ILL GIVE THE BEST ANSWER BRAINLIST

Answer:

C. 2/5 full

Step-by-step explanation:

[tex]\frac{1}{\frac{1}{15} } :\frac{6}{y}[/tex]

y · 1 = 1/15 · 6

y = 0.4

[tex]y=\frac{2}{5}[/tex]

Answer:

third option mechanical weathering

Step-by-step explanation:

in mechanical weathering it breaks the rock into smaller pieces . these smaller pieces are just like the bigger rock , just smaller .

Answer:

x=3          x=-2

Step-by-step explanation:

(x-3)(x+2)=0

Using the zero product property

x-3 =0   x+2 =0

x=3          x=-2

Answer:

3 and -2

Step-by-step explanation:

x -3 =0

x= 3

x+2 =0

x= -2

Answer:

Ok

Step-by-step explanation:

So,

Answer:

I think the answer may be 465

Step-by-step explanation:

105+115=220

150+95=245

245+220=465

Question:

5/6 x 3/4 as an array is?

Answer:

0.625 or 5/8

Step-by-step explanation:

Using Arrays to Explore Numbers

Arrays are useful models for multiplication which can be used in a variety of ways, ranging from highly structured lessons to games and open investigations.

An array is formed by arranging a set of objects into rows and columns. Each column must contain the same number of objects as the other columns, and each row must have the same number as the other rows.

The following array, consisting of four columns and three rows, could be used to represent the number sentence 3 x 4 = 12, 4 x 3 =12, 3 + 3 + 3 + 3 = 12 and 4 + 4 + 4 =12.

Building Multiplication Facts and Tables

Arrays can be used for building multiplication facts in a meaningful way. Before drilling and memorising tables, children must understand how these facts are derived. For example, by progressively adding another column of three objects, children can build the three-times tables for themselves. This representation not only assists in understanding the process, but provides a visual image for children to draw upon as they begin to use and memorise the basic number facts.

Using arrays to explore larger numbers

Arrays can be helpfully used to explore calculations such as 13 x 5 where the array can be split into useful chunks such as 10 and 3. This means that children can use their known number facts to work out calculations.

Here 13 x 5 = (10 x 5) + (3 x 5).

After a while drawing all the dots can get very tedious! The blank array then becomes a very useful tool for helping children model their thinking and work out how to do more complex multiplications in an informal way.

Here's a child using the blank array, as a thinking tool, to help them work out 15 x 14.

The blank array helps children to use other strategies, such as compensating, when carrying out multiplication. Here, to work out 34 x 9, the child has decided to do 34 x 10 and then take off the 34 x 1.

Beyond the blank array this 'dividing the multiplication into easy parts' strategy can be formalised into the grid method. The children can see how the 'abstract' grid method overlays the array and formalises the blank array into a standard form.

Division as the Inverse Operation of Multiplication

Of the four operations, division is the most troublesome for young students. Full understanding of division tends to lag well behind the other operations. For many children opportunities to explore the concept with concrete materials are curtailed well before they perceive the relationships between division and the other three operations. One such relationship, the inverse relationship between division and multiplication, can be effectively illustrated using arrays.

For example; 3×5=15  or 3 rows of 5 make 15, can be represented by the following array.

Looking at the array differently reveals the inverse, that is

15÷3=5  or 15 put into 3 rows makes 5 columns - or 5 in each row.

Language clearly plays an important role in being able to express the mathematical relationships and the physical array supports this aspect of understanding by giving the children a concrete image to talk about.

Placing the mathematics into a real-life context through word problems can facilitate both understanding of the relationship and its expression through words.

For example, "The gardener planted 3 rows of 5 seeds. How many seeds did she plant?" poses quite a different problem to "The gardener planted 15 seeds in 3 equal rows. How many seeds in each row?" yet both these word problems can be modelled using the same array.

Further exploration of the array reveals two more ways of expressing inverse relationships: 5×3=15 and 15÷3=5 .

The word problems can be adapted to describe these operations and highlight the similarities and differences between the four expressions modelled by the one array.

Using the blank array

Suppose you want to figure out 176 ÷ 8. We can set this up as an array with the value of one side missing.

Using known multiplication facts the value of the missing side can be built up.

So the child can see that 22 lots of 8 is the same as 176.

The array is a very powerful tool for supporting the development of children's thinking around both multiplication and division.

Answer:

To find the surface area of a cuboid, add the areas of all 6 faces. We can also label the length (l), width (w), and height (h) of the prism and use the formula, SA=2lw+2lh+2hw, to find the surface area.

Step-by-step explanation:

SA=2*4*5+2*4*6+2*6*5

=148 cmStep-by-step explanation:

Answer:

perpendicular, parallel, neither, parallel

Answer:

v<31

Step-by-step explanation:

Simplify

v-6<25

We want to isolate x.

v<25+6

Simplify

v<31

If this helps please mark as brainliest

Answer:

using pythagoras theorem

h^2=p^2+b^2

h^2=12^2+9^2

h^2=144+81

h^2=225

h=[tex]\sqrt{225}[/tex]

h=15

Step-by-step explanation:

Answer:

Graph A

Step-by-step explanation:

Graph A flour is half the sugar

Graph B folur is twice the sugar

Graph C flour is equal the sugar

Answer:

7

Step-by-step explanation:

210=5×6×x

x is length of aquarium.

210÷(5×6)=7

$/sqrt{9}$

Answer:

1.333333333333 or 4/3

Happy Pride Month!

Part 1:

You can solve the pair of equation graphically by writing the y-intercepts which are (0,-1) and (0,-5) Then follow the slopes of each line till the intercept. The slopes are up 2 over 1 and up 4 over 1 respectivily

Part 2:

When you graph these you get the answer of (2,3)

x is 2 and y is 3

Have a good day!

Answer:

Step-by-step explanation:

Answer:

http://web.mnstate.edu/peil/MDEV102/U4/S36/S363.htmlStep-by-step explanation:

Answer:

First quartile: 35

Third Quartile: 46

Step-by-step explanation:

Answer:

[tex]1017.9ft ^{3} [/tex]

Step-by-step explanation:

[tex]volume \: of \: cylinder \\ v = \pi \times r {}^{2} \times h \\ = \pi \times 9^2 \times 4 \\ = 1017.9[/tex]

[tex]4.5[/tex] ✅

Step-by-step explanation:

[tex] \frac{1}{2} \times 9 \\ = \frac{1 \times 9}{2} \\ = \frac{9}{2} \\ = 4.5[/tex]

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Answer:

4.2

Step-by-step explanation:

If they're similar, the fraction between their sides need to be equal, so do this:

2/1.4 = 6/MN

6•1.4/2 = MN

MN = 4.2

Given:

Length of arc AB is 4 units.

The measure of central angle corresponding to arc AB is 30 degrees.

To find:

The circumference of the circle.

Solution:

The formula for arc length is:

[tex]s=2\pi r\dfrac{\theta }{360^\circ}[/tex]

Where, r is the radius and [tex]\theta[/tex] is the central angle.

It is also written as:

[tex]s=C\dfrac{\theta }{360^\circ}[/tex]

Where, C is the circumference of the circle.

Putting [tex]s=4,\theta =30^\circ[/tex] in the above formula, we get

[tex]4=C\dfrac{30^\circ}{360^\circ}[/tex]

[tex]4=\dfrac{1}{12}C[/tex]

Multiply both sides by 12.

[tex]4\times 12=C[/tex]

[tex]48=C[/tex]

The circumference of the circle is 48 units.

Therefore, the correct option is A.

Answer:

D. 49

Step-by-step explanation:

Create a proportion where x is the number of liters of yellow paint she should use:

[tex]\frac{4}{7}[/tex] = [tex]\frac{28}{x}[/tex]

Cross multiply and solve for x:

4x = 196

x = 49

So, she should add 49 liters of yellow paint.

The correct answer is D. 49

Answer:

13

Step-by-step explanation:

Answer:

13

Step-by-step explanation:

48-[(2^2+3)x5]

48-[(4+3)x5]

48-[(7)x5]

48-[35]

13

Given :

A motorboat travels 9 miles downstream (with the current) in 30 minutes.

The return trip upstream (against the current) takes 90 minutes.

To Find :

The speed of stream and motorboat.

Solution :

Let, speed of stream is u and speed of motorboat is v.

Now, we know distance is same in both case.

Distance( downstream ) = speed × time

[tex]9 = ( v +u ) \times 0.5\\\\v + u = 18[/tex]....1)

Distance( upstream ) = speed × time

[tex]9 = ( v- u ) \times 1.5\\\\v-u = 6[/tex]....2)

Solving equation 1 and 2 , we get :

v = 12 miles/hour and u = 6 miles/hour

Hence, this is the required solution.

Answer:

The speed of the stream is 6 mph, and the speed of the boat is 12 mph.

Step-by-step explanation:

Downstream:

9 miles in 30 minutes (0.5 hour).

9 mi / 0.5 h = 18 mph

Upstream:

9 miles in 90 minutes (1.5 h).

9 mi / 1.5 h = 6 mph

Let s = speed of stream.

The speed of the stream adds to the speed of the boat downstream and is subtracted fromt he speed of the boat upstream.

Let speed of boat be b.

b + s = 18

b - s = 6

Add the equations.

2b = 24

b = 12

b + s = 18

12 + s = 18

s = 6

The speed of the stream is 6 mph, and the speed of the boat is 12 mph.

Given:

The population of animals = 350

Growth rate = 5% each year.

To find:

The exponential function to model this situation.

Solution:

The general exponential growth model is:

[tex]f(t)=a(1+r)^t[/tex]

Where, a is the initial value, r is the growth rate in decimal and t is the time period.

Substituting [tex]a=350, r=0.05[/tex] in the above formula, we get

[tex]f(t)=350(1+0.05)^t[/tex]

[tex]f(t)=350(1.05)^t[/tex]

Therefore, the required exponential function is [tex]f(t)=350(1.05)^t[/tex], where f(t) is the number of animals after t years.

Answer:

The initial height of the ball before launch is 393.12 feet.

Step-by-step explanation:

The ball's height ([tex]h[/tex]), in feet, as a function of time ([tex]t[/tex]), in seconds, is represented by [tex]h(t) = -16\cdot t^{2}+153.6\cdot t + 393.12[/tex]. The initial height of the ball before launch is the height of the ball evaluated at [tex]t = 0\,s[/tex]:

[tex]h(0) = -16\cdot (0)^{2}+153.6\cdot (0) + 393.12[/tex]

[tex]h(0) = 393.12\,ft[/tex]

The initial height of the ball before launch is 393.12 feet.

Answer:

a. 10.08

b. 0.0576

Answer:

m`/_DAB` will be equal to m`/_CAB`.

Step-by-step explanation:

Correct on edmentum.