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2 years ago

Write an algebraic expression to represent each scenario. Part A: “the sum of 6 and y”

Mathematics

1 answer:

Nostrana [21]2 years ago

3 0

Answer:

6 + y

Step-by-step explanation:

Here, we want to write an algebraic expression to represent the situation

Mathematically, we want to add 6 and y

The algebraic expression will be;

6 + y

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Write an inequality to model the situation. a number exceeds 9.

sveta [45]

n > 9 would be the answer

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50 POINTS! Could someone answer these 3 questions? <br><br> I’ll mark Brainliest

evablogger [386]

Answer:

Where's the chart

Step-by-step explanation:

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3 years ago

The figure shows △ABC . BD is the angle bisector of ∠ABC .

Misha Larkins [42]

From angle bisector theorem, we know that:

$\dfrac{|AD|}{|DC|}=\dfrac{|BA|}{|BC|}\\\\\\\dfrac{|AD|}{|DC|}=\dfrac{8}{10}=\dfrac{4}{5}$

Moreover:

$|AD|+|DC|=6$

so:

$|DC|=6-|AD|$

and

$\dfrac{|AD|}{|DC|}=\dfrac{4}{5}\\\\\\\dfrac{|AD|}{6-|AD|}=\dfrac{4}{5}\qquad\qquad\text{cross multiplying}\\\\\\ 5\cdot|AD|=4\cdot\big(6-|AD|\big)\\\\5\cdot|AD|=24-4\cdot|AD|\\\\5\cdot|AD|+4\cdot|AD|=24\\\\9\cdot|AD|=24\qquad|:9\\\\\\|AD|=\dfrac{24}{9}=\boxed{\dfrac{8}{3}}$

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3 years ago

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Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

$\to x$ )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence $a_n$ such that:

$a_n\ belongs\ to\ Q$

and

$a_n\to \alpha$

$f(a_n)=ka_n$

( since $a_n$ belongs to Q )

Let f is continuous at x=α

This means that:

$f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha$

This means that:

$f(\alpha)=k\alpha$

This means that:

f(x)=kx for every x∈ R

4 0

2 years ago

Can someone solve this! Need it done soon

Airida [17]

Answer:

117.5 ft²

Step-by-step explanation:

To find the area of the shaded area, we can find the area of the square, then subtract the areas of the two semicircles.

First, we will find the area of the given square, by using the formula $lw$ , multiplying the length and width. The dimensions of this square are 14×14.

14 · 14 = 196

The area of the square is 196 ft².

We can now find the area of the two congruent half-circles. Since they are identical, we can simply find the area of one circle if it was whole. To find the area of a circle, we'll use the formula $A=\pi r^2$ . With some simple deduction, we can see that the diameter of the circle is 10 ft, so the radius would be 5 ft long. Plug our values into the formula.

A = $\pi$ 5²

We will use 3.14 for $\pi$ .

A = 78.5

The area of both the semicircles is 78.5 ft².

Now, we can subtract.

196 - 78.5 = 117.5

The area of the figure is 117.5 ft².

Good luck ^^

7 0

2 years ago