Ask question

Ask question

All categories

valentina_108 [34]

3 years ago

7

Compare the ratios in Table 1 and Table 2. Table 1 3 5 6 10 9 15 12 20 Table 2 7 10 14 20 21 30 28 40 Which statements about the

ratios are true? Check all that apply

1 answer:

Hatshy [7]3 years ago

4 0

Answer:

pp

Step-by-step explanation:

You might be interested in

Plzz help will mark brainleast

iris [78.8K]

Answer:

A-1 central nervous system

b- peripheral nervous

5 0

3 years ago

A rectangle has perimeter, P, length, land width, w. Which of the following represents lin

almond37 [142]

Answer:

$l=\frac{P}{2}-w$ .

Step-by-step explanation:

The perimeter of a rectangle is the sum of it's side lengths.

A rectangle has 4 sides where it's opposite sides are congruent.

So if one side has measurement w, then there is another side that has measurement w.

If there is one side that has measurement l, then there is another side that has measurement l.

So if you add w+w+l+l you get 2w+2l.

They are giving us that the perimeter is P, so P=2w+2l.

we are being asking to solve for l.

P=2w+2l

First step: Isolate term that contains the l, so get 2l by itself first.

We are going to subtract 2w on both sides giving us:

P-2w=2l

2l=P-2w

Now that we have 2l by itself it is time to perform the last step in getting l by itself.

Second step: Divide both sides by 2.

This gives us:

l=(P-2w)/2

You may separate the fraction like so:

$l=\frac{P-2w}{2}=\frac{P}{2}-\frac{2w}{2}=\frac{P}{2}-w$ .

I don't know your options but I have solve for l in terms of P and w

and got $l=\frac{P}{2}-w$ .

Please let me know if you have further questions with this problem.

4 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

3 years ago

Which statement is true about the dependent and independent variables?

Hatshy [7]

Answer:

1. The independent variable goes on the x-axis and the dependent variable goes on the y-axis.

7 0

3 years ago

Factorise 2x^2 - 3x -2 < 0

Galina-37 [17]

Step 1: Factor $2 x^{2} - 3x -2$
1. <span> Multiply 2 by -2, which is -4.</span>
2. <span>Ask: Which two numbers add up to -3 and multiply to -4?
</span>3. <span>Answer: 1 and -4
</span>4. Rewrite $-3x$ as the sum of $x$ and $-4x$

$2 x^{2} +x-4x-2\ \textless \ 0$

Step 2: <span>Factor out common terms in the first two terms, then in the last two terms.

</span> $x(2x+1)-2(2x+1)\ \textless \ 0$
<span>
Step 3: </span>Factor out the common term $2x+1$

$(2x+1)(x-2)\ \textless \ 0$

Step 4: Solve for $x$

1. Ask: When will $(2x+1)(x-2)$ equal zero?
2. Answer: When $2x + 1 = 0$ or $x-2=0$
3. <span>Solve each of the 2 equations above:

</span> $x=- \frac{1}{2} ,2$
<span>
Step 5: </span>From the values of $x$ <span>above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2

Step 6: P</span><span>ick a test point for each interval

</span>For the interval $x\ \textless \ - \frac{1}{2} 
$
Lets pick $x=-1.$ Then, $2(-1) ^{2} -3 * -1 -2 \ \textless \ 0$
After simplifying, we get $3\ \textless \ 0$ , Which is false.
Drop this interval.
<span>
For this interval $- \frac{1}{2} \ \textless \ x\ \textless \ 2$
Lets pick $x=0$ . Then, $2* 0^{2} - 3 * 0-2\ \textless \ 0$ . After simplifying, we get $-2\ \textless \ 0,$ which is true. Keep this <span>interval.

For the interval </span> $x\ \textgreater \ 2$

Lets pick $x = 3.$ Then, $2 * 3 ^{2} -3*3-2\ \textless \ 0.$ After simplifying, we get $7\ \textless \ 0$ , Which is false. Drop this interval.

.Step 7: Therefore,
$- \frac{1}{2} \ \textless \ x\ \textless \ 2$

Done! :)</span>

4 0

4 years ago