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3 years ago

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If a system including the quadratic equation representing the parabola and a linear equation has no solution, which linear equat

ion could be the second equation in the system? A. 1/2x=y+4 B. 2x-y=0 C.y=6 D.y=2x+6

2 answers:

marshall27 [118]3 years ago

7 0

9514 1404 393

Answer:

D. y=2x+6

Step-by-step explanation:

The line cannot intersect the parabola if it has a y-intercept greater than 5 and a suitable slope. The only sensible answer choice is ...

y = 2x +6

Leni [432]3 years ago

3 0

Answer:

D. y = 2x + 6.

Step-by-step explanation:

The required equation would not intersect the parabola at any point.

The only one to fit that is D.

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If y is proportional to x and y = -15 when x = -20, what is the value of x when y = 12?

alexandr402 [8]

Maybe i have that you wanted We can represent the situation like that: Y=-15 and x=-20 Then When y=12 x=? We make x(The unknown) *-15=12*(-20) We have an simple équation to solve Thus x=(12*(-20))/-15 X=16

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3 years ago

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-2x-9y=-25 4x+9y=23<br><br><br><br> Help me

Tatiana [17]

X=25/11 and y=17/11 if that’s what your asking for

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2 years ago

Chandler earns a base pay of $2,200 per month. He also earns a commission of 4% of his total sales . One month Chander earned $2

Kipish [7]

The answer is $333.33 hope it helps

8 0

3 years ago

University Theater sold 465 tickets for a play. Tickets cost $23 per adult and $13 per senior citizen. If total receipts were $7

liq [111]

<u>Answer:</u>

355 senior citizen tickets were sold.

<u>Step-by-step explanation:</u>

Assuming a to the the ticket of adults and s to be the ticket of senior citizens, we can write two equations as:

$a +s=465$ --- (1)

$23a+13s=7145$ --- (2)

From equation 1, $a=465-s$ .

Substitute $a$ in equation 2 to get:

$23 (465-s) + 13s = 7145$

$10695 - 23s +13s =7145$

$-10s = -3550$

$s = 355$

Therefore, there were 355 senior citizen tickets sold.

8 0

3 years ago

Please help very urgent

Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by $\sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}$ .

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

$a_{n} = a_{1} • r^{n-1}$

Where $a_{n}$ is the nth term, $a_{1}$ is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, $a_{1}$ will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

$\sqrt[5]{\frac{a_{6}}{a_{1}}}$ =

$\sqrt[5]{\frac{1}{32}}$ =

$\frac{1}{2}$ .

Therefore: $r = \frac{1}{2}$ .

Using all the information we have, we can find the explicit rule:

$a_{n} = a_{1} • r^{n-1}$

$a_{1}$ = 32
$r = \frac{1}{2}$

$a_{n} = a_{1} • r^{n-1}$ →

$\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}$

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

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3 years ago

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