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ki77a [65]
3 years ago
9

Select whether the relationship between each pair of quantities is proportional.

Mathematics
1 answer:
vfiekz [6]3 years ago
7 0
2 is a great answer you should use......
You might be interested in
(2 tens 1 one)x10 in unit form and standard form
omeli [17]

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

<h3>How to write the number in standard form?</h3>

The number expression is given as:

(2tens 1 one) x 10

2 tens is represented as:

2 * 10

1 one is represented as:

1 * 1

So, the number expression can be rewritten as:

(2tens 1 one) x 10 = (2 * 10 + 1 * 1) x 10

Evaluate the product

(2tens 1 one) x 10 = (20 + 1) x 10

Evaluate the sum

(2tens 1 one) x 10 = (21) x 10

Evaluate the product

(2tens 1 one) x 10 = 210

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

Using the above steps as a guide, we have:

  • (5 hundreds 5 tens) * 10 ⇒ 5 thousands and 5 hundreds ⇒ 5500
  • (2 thousands 7 tens) / 10 ⇒ 2 hundreds and 7 units ⇒ 207
  • (4 ten thousands 8 hundred) / 10 ⇒ 4 thousands and 8 tens ⇒ 4080

Read more about standard form at

brainly.com/question/19169731

#SPJ1

4 0
2 years ago
One package of socks cost $7. How many packages can you but with $56?
Akimi4 [234]

Step-by-step explanation:

$7 = 1 package socks

%56 = ??

$56/$7 × 1 package socks

= 8 × 1

= 8 package socks

Hope it helps!

4 0
3 years ago
I rlly rlly need the answer ASAP
Natalija [7]

Answer:

your answer is c

Step-by-step explanation:

In the fraction Jonah made split his table into 10 parts to show 3/5 and to get the 10 parts he multiplied 5 by 2 to get 10 and you have to do the same to the 3 so 3 multiplied by 2 is 6

3 0
3 years ago
How do i solve this problem
Vladimir79 [104]
Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)

Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)

Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer:  {r = -17/26
               {s = 23/13                        {t = 21/26
7 0
3 years ago
Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3
chubhunter [2.5K]

Answer:

The simplified form is \dfrac{2(x-1)}{3(x+1)}.

x =1 is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

\dfrac{2a^2-4a+2}{3a^2-3}

Let us simplify the numerator and denominator separately.

The numerator is given as 2a^2-4a+2

2 is a common factor in all the three terms. So, we factor it out. This gives,

=2(a^2-2a+1)

Now, a^2-2a+1=(a-1)(a-1)

Therefore, the numerator becomes 2(a-1)(a-1)

The denominator is given as: 3a^2-3

Factoring out 3, we get

3(a^2-1)

Now, a^2-1 is of the form a^2-b^2=(a-b)(a+b)

So, a^2-1=(a-1)(a+1)

Therefore, the denominator becomes 3(a-1)(a+1)

Now, the given expression is simplified to:

\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}

There is (x-1) in the numerator and denominator. We can cancel them only if x\ne1 as for x=1, the given expression is undefined.

Now, cancelling the like terms considering x\ne1, we get:

\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}

Therefore, the simplified form is \dfrac{2(x-1)}{3(x+1)}

The simplification is true only if  x\ne1. So, x =1 is the excluded value for the given expression.

8 0
3 years ago
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