Ask question

Ask question

All categories

3 years ago

6

The equation y = 17x shows that a boat travels at a speed of 17 km/h. Which ordered pair satisfies the equation y = 17x? A. (0,

0) B. (0, 17) C. (1, 18) D. (17, 1) PLEASE HELP ASAP!

1 answer:

aleksklad [387]3 years ago

6 0

A, (0,0) is the answer

You might be interested in

What is the Y - Intercept?

schepotkina [342]

Y intercept is 24. ( y intercept is when x=0 )

3 0

3 years ago

Sarah is a computer engineer and manager and works for a software company. She receives a

daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = $a$

Given that:

$a_4=129\\a_{10}=207$

Formula for $n^{th}$ term of an Arithmetic Progression is given as:

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

Where $d$ will represent the number of projects increased every year.

and $n$ is the year number.

$a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)$

Subtracting (2) from (1):

$78 = 6d\\\Rightarrow d =13$

By equation (1):

$129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90$

<em>Number of projects in the first year = 90</em>

<em></em>

<em>b) </em>

Number of projects in the twelfth year =

$a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233$

Each project pays $500

Earnings in the twelfth year = 233 $\times$ 500 = $116500

Sum of an AP is given as:

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938$

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 $\times$ 1938 = $969000

8 0

2 years ago

Give the equation of the circle centered at the origin and passing through the point (0,-8)

Norma-Jean [14]

Answer:

The answer to your question is x ² + y² = 64

Step-by-step explanation:

Process

1.- Find the length of the radius

C (0, 0)

P (0, -8)

d = $\sqrt{(x2 - x1)^{2}+ (y2 - y1)^{2} }$

d = $\sqrt{(0- 0)^{2}+ (-8 - 0)^{2} }$

d = $\sqrt{64}$

d = 8

2.- Find the equation of the circle

$(x - h)^{2} + (y - k)^{2} = r^{2} \\$

h = 0 and k = 0

$(x - 0)^{2} + (y - 0)^{2} = 8^{2}$

x² + y² = 64

5 0

3 years ago

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

5 0

3 years ago

4.5w = 5.1w - 30<br><br> Please help!!!

Lelu [443]

4.5w = 5.1w - 30

4.5w - 5.1w = -30

-0.6w =-30

Divide both sides by -0.6

-0.6w/-0.6 = -30/-0.6

<span>w = 50

</span>I hope this helps.

6 0

3 years ago