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3 years ago

14

The equation y = 0.75x + 10 describes the cost, y, of placing an ad with x words in the classified section of a newspaper. How m

any words are in an ad that costs $25?

1 answer:

poizon [28]3 years ago

6 0

y = cost = $25

25 = 0.75 +10

subtract 10 from each side

15=0.75x

x = 15/0.75 = 20

there were 20 words in the ad

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A person can see the top of a building at an angle of 65°. The person is standing 50 ft away from

zhannawk [14.2K]

9514 1404 393

Answer:

(b) 112.2 ft

Step-by-step explanation:

The relevant trig relation is ...

Tan = Opposite/Adjacent

For the given geometry, this becomes ...

tan(65°) = (height above eye level)/(50 ft)

Then we have ...

(height above eye level) = (50 ft)tan(65°) = 107.2 ft

Adding the height of eye level will give us the height of the building.

building height = (eye level height) + (height above eye level)

building height = (5 ft) + (107.2 ft)

building height = 112.2 ft

6 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

Number 2 please thanks

Alex

Answer:

c) 2x+1

Step-by-step explanation:

Let's x be number of Nick's goals.

Emma has scored 1 more than twice the number of Nicks.

1 more is +1, of twices the number is 2x.

so 2x+1=15

6 0

3 years ago

Read 2 more answers

How do I do problem twelve you don’t have to give the answer because you can’t see the numbers but explain how I would do It

liraira [26]

In order to do this you would multiply how much she used for 1/16 of the garden by 16. So 1/2 • 16 which is 8. So the answer is 8.

Hope this helps!

8 0

3 years ago

Find the value of x.

Cerrena [4.2K]

$x^2 + (x + 3)^2 = (\sqrt{117})^2$

$x^2 + x^2 + 6x + 9 = 117$

$2x^2 + 6x -108 = 0$

$x^2 + 3x - 54 = 0$

$(x + 9)(x - 6) = 0$

$x + 9 = 0 ~~or~~ x - 6 = 0$

$x = -9 ~~ or~~ x = 6$

Since the side of a triangle cannot have a negative length, we discard the solution x = -9.

Answer: x = 6

3 0

3 years ago

Read 2 more answers