Ask question

Ask question

All categories

2 years ago

8

Donna wants to save $700 to buy a TV. She saves $17 cach week. The amount, A (in dollars), that she still needs after w weeks is

given by the following
function
A(w)= 700 - 17w
Answer the following questions.
(a) How much money does Donna still need after 6 weeks?
s
(b) If Donna still needs $411, how many weeks has she been saving?

1 answer:

alexandr1967 [171]2 years ago

8 0

Answer:

Part A:

She would need $598 after 6 weeks of collecting money.

Part B:

She would have waited 17 weeks to have $411 left.

Step-by-step explanation:

Part A:

Replace w in 700 - 17w with 700 - 17(6).

This would give you $598 left until $700 has been reached.

Part B:

Replace w in 700 - 17w with 700 - 17(17).

This would give you $411 left until $700 has been reached.

You might be interested in

What is the greatest common factor of 24s^3 , 12s^4 , and 18s

denpristay [2]

6s. 2 and 3 divide into all these numbers but six is the greatest number that all
Divide into and they all have one s at least.

6 0

2 years ago

Read 2 more answers

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

2 years ago

Read 2 more answers

Solve each inequality, graph the solution on the number line, and write the solution in interval notation

Lubov Fominskaja [6]

Answer:

$(-\infty,+\infty)$

Step-by-step explanation:

The given inequality is 3x-2>4 or $5x-3\leq 7$

We group similar terms to obtain: 3x>4+2 or $5x\le7+3$

Simplify to get:

3x>6 or $5x\le10$

x>2 or $x\le 2$

This implies that the solution set is all real numbers.

The solution in interval notation is $(-\infty,+\infty)$

5 0

3 years ago

a theater is designed with 15 seats in the first row, 19 in the second, 23 in the third, and so on. if this seating pattern cont

marusya05 [52]

Answer:

<u>131 seats</u> are in the 30th row.

Step-by-step explanation:

The theater is designed with the first row there are 15 seats, in second row 19 seats and in the third row there are 23 seats.

Now, to find the number of seats in the 30th row.

So, we get the common difference( $d$ ) from the arithmetic sequence first:

$19-15=4.$

Thus, $d=4.$

So, the first tem $a(1)$ = 15.

The number of last row ( $n$ ) = 30.

Now, to get the number of seat in the 30th row we put formula:

$a(n) = a(1) + d(n-1)$

$a(30)=15+4(30-1)$

$a(30)=15+4\times 29$

$a(30)=15+116$

$a(30)=131$

Therefore, 131 seats are in the 30th row.

4 0

3 years ago

The first term of an arithmetic sequence is -5, and the tenth term is 13. Find the common difference.

il63 [147K]

Answer:

2

Step-by-step explanation:

If the common difference is d we can write:

-5 + 9d = 13

9d = 18

d = 2

3 0

3 years ago

Read 2 more answers